Sum of Stresses in a Control Volume along $x$-axis

1. Your last line has some errors and should read (with two denominators changed)

$$=left(frac{partialsigma_{xx}}{partial x}+frac{partialtau_{yx}}{partial y}+frac{partialtau_{zx}}{partial z}right)dx,dy,dz$$

2. By "there opposite terms", I think you're asking why $[sigma_{xx}(dy,dz)+tau_{xy}(dx,dz)+tau_{zx}(dx,dy)]$ is subtracted. The reason is that we're performing a force balance on the element in all three directions to obtain the net force in the $x$ direction $F_{rm{stress,}x}$. We're also applying a Taylor series expansion to the stress, which is assumed to vary inside the element.

   Take the stress $sigma_{xx}$, for example. On the left side of the element, call it $sigma_{xx}(text{at }x=0)$. On the right side, call it $sigma_{xx}(text{at }x=dx)$. So we have a force acting to the left with magnitude $$sigma_{xx}(text{at }x=0),dy,dz,$$ and a force acting to the right with magnitude $$sigma_{xx}(text{at }x=dx),dy,dz.$$ Equilibrium in the $x$-direction (taking rightward forces as positive) thus gives $$-left[sigma_{xx}(text{at }x=0),dy,dzright]+left[sigma_{xx}(text{at }x=dx),dy,dzright]=0,$$ or, with the Taylor expansion, $$-left[sigma_{xx}(text{at }x=0),dy,dzright]+left[sigma_{xx}(text{at }x=0),dy,dz+frac{partial sigma_{xx}(text{at }x=0)}{partial x}dxright]$$ $$=frac{partial sigma_{xx}(text{at }x=0)}{partial x}dx,$$ or $frac{partial sigma_{xx}}{partial x}dx$ for simplicity, which is one component of the expression above. The same process is applied to the $tau_{xy}$ and $tau_{xz}$ stresses, each of which also applies an $x$-direction force. Does this make sense?

   (If it helps, if there's something that's not clicking, the same derivation appears here and here.)