How can I plot the direction field for a differential equation?

If you wish to explore the solutions to an equation I'd suggest the EquationTrekker package. Have a look at the documentation.

Needs["EquationTrekker`"]
EquationTrekker[y'[x] == (y[x] + Exp[x])/(x + Exp[y[x]]), y, {x, -5, 5}]

Here is something you can do quickly and gives a nice understanding of the behavior for positive and negative initial conditions:

s[r_?NumericQ] := 
               NDSolve[{D[y[x], x] == (y[x] + E^x)/(x + E^y[x]), y[0] == r}, y, {x, 0, 5}]

Plot[Evaluate[
  y[x] /. s[#] & /@ Union[Range[-2, 0, .1], Range[-.10, 10, 1]]], {x, 0, 15}, 
  PlotRange -> Full]

No need to solve the differential equation to generate a direction field. According to the Wikipedia lemma on slope fields you can plot the vector {1, (y + Exp[x])/(x + Exp[y])}:

VectorPlot[{1, (y + Exp[x])/(x + Exp[y])}, {x, 0, 2}, {y, 0, 2}]

or perhaps you could use a stream plot:

StreamPlot[{1, (y + Exp[x])/(x + Exp[y])}, {x, 0, 2}, {y, 0, 2}]

I approached this slightly differently, in an attempt to mimic what's written in a textbook of mine for direction fields. I set the plot as a system of unit vectors, with x and y from this derivation:

$$mathit{x}^2+mathit{y}^2=1wedge mathit{m}=frac{y}{mathit{x}}Rightarrow mathit{x}=frac{mathit{y}}{mathit{m}}Rightarrow frac{mathit{y}^2}{mathit{m}^2}+mathit{y}^2=1Rightarrow mathit{y}=frac{mathit{m}}{sqrt{mathit{m}^2+1}} $$
$$ mathit{y}=mathit{m} mathit{x}Rightarrow mathit{m}^2 mathit{x}^2+mathit{x}^2=1Rightarrow mathit{x}=frac{1}{sqrt{mathit{m}^2+1}} $$
So I put this into Mathematica

func =
  Function[{m},
    VectorPlot[
      {1/Sqrt[1 + m[x, y]^2], m[x, y]/Sqrt[1 + m[x, y]^2]}, 
      {x, -4, 4}, {y, -4, 4},
      VectorPoints -> Fine]];

When given the original equation and one of the ones from my textbook it displays:

Column[func /@ {Function[{x, y}, (E^x + y)/(E^y + x)], Function[{x, y}, y (y - 3)]}]

Which gets very close to what I see my textbook. The range and options can be adjusted as needed.

In the most popular contemporary undergraduate calculus textbooks, including those by Larson and Edwards, Stewart, Rogawski and Adams, and others, a slope field (also called a direction field) is a plot of short line segments at grid points all having the same length and without an arrowhead indicating direction. A slope field indicates only the slope of the solution curve at each grid point by the slope of the line segment only. Only Wesley Wolfe's answer approaches this method of plotting slope fields as of this writing. I adjust a couple of options to VectorPlot[] in his answer below.

We assume that the D.E. can be written in the form $y'=F(x,y)$, where $F(x, y)$ is a function only of two variables $x$ and $y$. Then we can plot the slope field in Mathematica as follows:

F[x_, y_] := (E^x + y)/(E^y + x)

VectorPlot[
  {1, F[x, y]}/Sqrt[1 + F[x, y]^2],
  {x, -4, 4}, {y, -4, 4}, 
  VectorScale -> 0.02, 
  VectorPoints -> Fine,
  VectorStyle -> "Segment"
]

There are some textbooks that prefer to put arrows on the line segments. To emulate this effect, change VectorStyle from "Segment" to "Arrow". However, this style is misleading and should be avoided, as the arrows generally do not point in the direction of the solution curves as they are traced out. To take an example from my bookshelf, Boyce and DiPrima's Differential Equations and Boundary Value Problems, 7th edition., p. 40-41: We consider $y'=frac{x^2}{1-y^2}$ which has general solution $-x^3+3y-y^3=c$. They give the slope field—with arrowheads—in figure 2.2.1, but you can plot it with Mathematica as I have along with a few solution curves below:

F[x_, y_] := x^2/(1 - y^2);

Show[ VectorPlot[
  {1, F[x, y]}/Sqrt[1 + F[x, y]^2],
  {x, -4, 4}, {y, -4, 4}, VectorScale -> 0.03, VectorPoints -> Fine, 
  VectorStyle -> "Arrow"], 
 Table[ContourPlot[-x^3 + 3 y - y^3 == c, {x, -4,  4}, {y, -4, 4}, 
   ContourStyle -> Red], {c, {-4, -1, 1,  4}}]]

At some point somebody must have objected to this abomination, because the arrowheads are gone from all slope fields in the 9th edition!

In Mathematica 12, VectorPlot is much more user-friendly. You don't need to mess with VectorScale parameters. The default behavior of VectorPlot is to draw direction-field-style vectors. https://reference.wolfram.com/language/ref/VectorPlot.html

Here is how I define my direction field plot function:

DirectionFieldPlot[F_, (* expression for y' *)
  xr_, (* x range *)
  yr_, (* y range *)
  init_, (* initial conditions (list of points) *)
  colors_List : {}, (* colors corresponding to initial points *)
  OptionsPattern[{LineWidth -> Thick, AspectRatio -> 1, 
    VectorPoints -> 30}] (* Linewith is thickness of the curves. 
  The other two are parameters of VectorPlot *)
  ] :=
 Module[{cv},
  cv = If[TrueQ[colors == {}], 
    Table[ColorData["Rainbow"][i], {i, 
      Range[0, 1, 1/(Length[init] + 1)][[2 ;; -2]]}], colors];
  Show[VectorPlot[{1, F}, xr, yr, 
    VectorPoints -> OptionValue[VectorPoints], 
    VectorStyle -> "Segment", Axes -> True, AxesOrigin -> {0, 0}, 
    VectorColorFunction -> None, 
    PlotRange -> {Take[xr, -2], Take[yr, -2]}, 
    AspectRatio -> OptionValue[AspectRatio]],
   StreamPlot[{1, F}, xr, yr, 
    PlotRange -> {Take[xr, -2], Take[yr, -2]}, 
    StreamPoints -> {Table[{init[[i]], {cv[[i]], 
         OptionValue[LineWidth]}}, {i, Length[init]}], Automatic, 
      10^50}, StreamScale -> None]]
  ]

To avoid DSolve and NDSolve, I use StreamPlot to plot the curves. You don't need to solve a differential equation for visualization.

Here are some examples:

DirectionFieldPlot[(y + E^x)/(x + E^y), {x, 0, 5}, {y, -2.5, 6}, Join[{{0, -1.5}, {0, -1.3}}, Table[{0, i}, {i, -1, 5, 1}]]]

DirectionFieldPlot[15 - 3 y, {x, 0, 3}, {y, 0, 6}, Table[{0, i}, {i, 0, 6, 1}], AspectRatio -> 1/2]

DirectionFieldPlot[x^2/y^2, {x, -3, 3}, {y, -3, 3}, Table[{0, i}, {i, -2.5, 2.5, 0.5}], VectorPoints -> 40]