I have the question:
$f(x,y) = y^2+x^2y+x^2-2y$
By setting the derivatives with respect to x and y both equal to $0$, I was able to find the critical points $(2,-1,3)$ and $(-2,-1,3)$. I simply found that y was equal to $-1$ by solving the derivative with respect to $x = 0$, then plugged $-1$ into the second equation to get $-2$ and $2$. However, there is a third critical point, $(0,1,-1)$, which I simply cannot discover a way to find using algebra. What mistake have I made?
You cannot divide by zero! Let us see what we have... We have the partial derivatives wrt $x$ and wrt $y$ both set to zero to form a system of equations:
$$ \begin{matrix} f_x=2x+2xy=0 \\ f_y=x^2+2y-2=0 \\ \end{matrix} $$
From this point you got $2xy=-2x$ and then $y=-2x/2x=-1$ but we cannot do the last step, because $x$ can be zero. The correct procedure is to take common factor: $2x(1+y)=0$ and see we have two possibilities, $x=0$ and $1+y=0$. The first one, plugged in the second equation completes the missing solution.