I would like some help sketching the critical points of a given function.
The function $f$ is given by $f (x, y) = (\sin x + \sin y)^2$ for all $(x, y)$ in $\mathbb R^2$.
The critical points we have gotten so far are: $(((π/2)+2πn),(π/2)+2πn)$, $((π/2)+ 2πn,(3π/2)+2πn)$, $((π/2) + πn, (π/2) + 2πn)$, $((3π/2) + 2πn, (π/2) + 2nπ)$, $((π/2)*πn, (3π/2)*2πn)$.
The conclusion I've made from this is that they either should be multiples of $π/2$, or be on the line $y = -x$.
Now I would want to get help drawing the critical points together with the level curve $ f(x,y) = 0 $.
Start by drawing the level curve $f(x,y) = \sin x + \sin y = 0$. Note that for a fixed $x_0$, to find the points $(x_0, y)$ that satisfy $f(x_0,y) = 0$, we need to solve $\sin y = -\sin x_0$. Thus,
$$ y = \arcsin( -\sin x_0) + 2\pi k = - \arcsin( \sin x_0 ) + 2\pi k = -x_0 + 2\pi k $$
or
$$ y = \pi - \arcsin( -\sin x_0) + 2\pi l = x_0 + \pi(2l + 1) $$
where $k, l \in \mathbb{Z}$.
Thus,
$$ f^{-1}(0) = \bigcup_{k \in \mathbb{Z}} \{ (x, -x + 2\pi k) \, | \, x \in \mathbb{R} \} \cup \bigcup_{l \in \mathbb{Z}} \{ (x, x + \pi(2l + 1) \, | \, x \in \mathbb{R} \} = \\ \bigcup_{k \in \mathbb{Z}} \{ y = -x + 2\pi k \} \cup \bigcup_{l \in \mathbb{Z}} \{ y = x + \pi (2l + 1) \} $$
is the union of infinitely many pairwise parallel lines $y = -x + 2\pi k$ together with infinitely many pairwise parallel lines $y = x + \pi (2l + 1)$ that intersect the lines $y = -x + 2\pi k$ at right angles. The union forms a grid depicted here that divides the plane into squares whose edges are of length $\sqrt{2} \pi$. Along the grid, the function has a (global) minimum.
Then, add to your picture the critical points $ \left( \frac{\pi}{2} + \pi k, \frac{\pi}{2} + \pi l \right)$ which are the global maxima of your function. Each point will be the middle point of each of the squares depicted above.