Determine whether or not it's true that "$f(x,y) = -(x^2 - 1)^2 -(x^2y -x -1)^2$ has only two critical points, and they're local maximum points"

Question

I believe this is not true because y is not independent from x on any of the first partial derivatives, therefore there are infinitely many critical points for x = 0. Is this correct?

Answer 1

Taking partial derivatives: $$\begin{cases}f_x=-2(x^2-1)\cdot 2x-2(x^2y-x-1)(2xy-1)=0 \\ f_y=-2(x^2y-x-1)\cdot x^2=0\end{cases} \Rightarrow \begin{cases} x_{1,2}=\pm 1 \\ y_{1,2}=2; 0\end{cases}.$$ The second derivative test: $$f_{xx}(1,2)=-26<0, f_{yy}(1,2)=-2<0, \Delta=(-26)(-2)-(-6)^2>0 \Rightarrow \text{max},$$ $$f_{xx}(-1,0)=-10<0, f_{yy}(-1,0)=-2<0, \Delta=(-10)(-2)-2^2>0 \Rightarrow \text{max}.$$

Answer 2

we have $$-(x^2-1)^2-(x^2y-x-1)^2\le 0$$ for all $x,y$ and this will be reached for $x=-1,y=0$