Q:

Local Max/Min of $ x^2y^2 - \log y$

It is asked to find the local maximum/minimum or saddle points of the function
$$f(x,y)=x^2y^2 - \log y$$
The answer says that $(0,1)$ is a saddle point and $(\frac{1}{2e^2}, e)$ is minimum. But the partial derivatives are
$$f'_x = 2xy^2 \text{ and } f'_y=2x^2y - \frac{1}{y}$$
The first one equals zero if $x=0 $ or $y=0.$ Since $y>0$, we should have that x=0, which doesnt satisfies $f'_y=0$
Am I doimg anything wrong or the answer is wrong? I dont see any extreme points.
Thanks!

A:

From your given answer, I suppose the function should be
$$f(x,y)=x^2y^2-x\log y,$$
where
$$f'_x=2xy^2-\log y,\quad f'_y=2x^2y-\frac{x}{y}.$$

