Suppose $f(x,y)=(x-y)(36-xy)$. Find the local maxima, minima, and saddle points. I currently know how to find $f_x$, $f_{xx}$, $f_y$, $f_{yy}$, and $f_{xy}$ and setting them equal to zero, but I'm confused how to tell the local maximum, minimum, or saddle points.
I have $f_x:y^2-2xy+36=0$, $f_y:-x^2+2xy-36=0$, $f_{xx}=-2y$, $f_{yy}=2x$, and $f_{xy}=2y-2x$.
From here I found $y^2-x^2=0$, so therefore $x=\pm y$ and when $x=y$, then $f_x=x^2-2x^2+36=0$ so $y=x=\pm 6$ which gives us the critical points (-6,-6),(6,6).
From here I have no idea what to do.

You can proceed as follows
Find the following values at the point $(6,6)$ as follows
$$r=\left(fxx\right)_{x=6, y=6}=-12, \ \ s=\left(fxy\right)_{x=6, y=6}=0, t=\left(fyy\right)_{x=6, y=6}=12$$
$$D=rt-s^2=(-12)(12)-0^2=-144$$
Since $D<0$ hence the function $f(x,y)$ has a saddle point $(6,6)$
Similarly, find the following values at the point $(-6,-6)$ as follows
$$r=\left(fxx\right)_{x=-6, y=-6}=12, \ \ s=\left(fxy\right)_{x=-6, y=-6}=0, t=\left(fyy\right)_{x=-6, y=-6}=-12$$
$$D=rt-s^2=(12)(-12)-0^2=-144$$
Since $D<0$ hence the function $f(x,y)$ has another saddle point $(-6, -6)$
Both the saddle points $(6,6)$ & $(-6,-6)$ are shown in plots below

