How to check if two functions only touch in one point?

Let's take "polar" coordinates for intersection curve projection: $$begin{array}{}x=rsqrt{cos phi} \ y = r sqrt{sin phi} end{array}$$ we will have $$x^4+y^4-xy-x^2=0 to r^2 = sqrt{sin phi cos phi} - cos phi$$ From where you see whole curve for intersection projection.

Suppose $(x,y)inmathbb{R}^2$ is a point other than $(0,0)$ satisfying the equation $$ x^4+y^4-xy=x^2 $$ Necessarily $xne 0$.

Letting $m={large{frac{y}{x}}}$, the equation reduces to $$ x^2(1+m^4)=1+m $$ hence for any $m > -1$ we get points $(x,y)$ given by $$ leftlbrace begin{align*} x&=pmsqrt{frac{1+m}{1+m^4}}\[4pt] y&=mx\[4pt] end{align*} right. $$ Note that allowing $m=-1$ in the above would yield $(x,y)=(0,0)$, hence the set of points $(x,y)inmathbb{R}^2$ satisfying the given equation is the set of points $(x,y)$ parameterized by $m$ as specified above, where $mge -1$.

In particular, there are infinitely many solutions.

But note:

The phrase "the points where two surfaces touch" is not always interpreted as meaning the same as "the points where two surfaces meet". Often it's intended to mean "the points where two surfaces meet and are mutually tangent (i.e., have the same tangent plane)".

With that interpretation, suppose $(a,b)$ is a point where for the two surfaces $$ leftlbrace begin{align*} z&=f(x,y)=x^4+y^4-xy\[4pt] z&=g(x,y)=x^2\[4pt] end{align*} right. $$ "touch".

By standard multivariable calculus methods

• The tangent plane at $(a,b)$ to the surface $z=f(x,y)$ has normal vector $langle{b-4a^3,a-4b^3,1}rangle$.$\[4pt]$
• The tangent plane at $(a,b)$ to the surface $z=g(x,y)$ has normal vector $langle{-2a,0,1}rangle$.

hence, in order for the surfaces to meet and the tangent planes to be the same, we must have $$ leftlbrace begin{align*} a^4+b^4-ab&=a^2\[4pt] b-4a^3&=-2a\[4pt] a-4b^3&=0\[4pt] end{align*} right. $$ If $b=0$ then $a=0$, which yields the solution $(a,b)=(0,0)$.

Suppose we have a solution $(a,b)$ with $bne 0$.

Replacing $a$ by $4b^3$, the above system reduces to $$ leftlbrace begin{align*} 256b^8-3&=16b^2\[4pt] -256b^8+1&=-8b^2\[4pt] end{align*} right. $$ which sums to $$8b^2=-2$$ contradiction, since $8b^2$ can't be negative.

Thus the only solution is $(a,b)=(0,0)$.

Thus while the point $(0,0)$ is one of infinitely many points where the two surfaces meet, it's the only point where the two surfaces "touch" (with the mutually tangent interpretation of "touch").

Well, the statement is not true... The picture below represents the set of points where both functions have the same value. It is a curve defined implicitly by $x^4+y^4-xy-x^2=0$.