Find the local maximum and minimum values and saddle point(s) of the function. If you have three-dimensional graphing software, graph the function with a domain and viewpoint that reveal all the important aspects of the function. (Enter your answers as a comma-separated list. If an answer does not exist, enter DNE.)
I found that there is no local minimum or maximum. However, the saddle point I found is $\{\frac{\pi}{2},0,0\},\{\frac{3\pi}{2},0,0\}$? but the answer is not correct.
Given $f(x,y) = 3y cos(x)$
No, this function has no maxima or minima. For $x = k \pi, k \in Z, cos (x) = \pm 1$ so as $ y \rightarrow \pm \infty , 3y cos(x) \rightarrow \pm \infty$ When both partial derivatives are zero, $\partial f / \partial x = -(3y) sin(x) = 0 $ so $y = 0$ or $x = k \pi$, AND $\partial f / \partial y = 3 cos(x) = 0$ so $x = (2k+1) \pi / 2$. The pairs with both derivatives zero are $ (x,y) = ((2k+1) \pi / 2, 0)$ (You had two of these points but there are infinite)
For a saddle point, you must have an upwards curvature in one direction and a downwards curvature in another non-parallel direction. $\partial^2 f / \partial x^2 = -(3y) cos(x) = 0 $ when $ y = 0$, so we have no curvature along the x direction. $\partial^2 f / \partial y^2 = 0 $ always; the "level curves" are straight lines so there is no curvature in the y direction. BUT there are other directions than parallel to the axes, and a zero second derivative test is "not enough info", rather than a definite fail.
However if you look at the graph you see something does happen at the points $ (x,y) = ((2k+1) \pi / 2, 0)$. If you travel along oblique lines such as $ y = \pm (x - (2k+1) \pi / 2 )$ you see increasing z (positive $ \partial f / \partial x$) on one of these lines, both sides of the given points, and negative on the other line. Work out the derivatives and check. So I find an infinity of saddle points.