Since the equation involves fractions, I want to clear the denominators by multiplying both sides by the least common multiple of $36$ and $16$, which is $144$. This gives me \[4x^2 + 9(y+5)^2 = 0\] Now I notice that both terms on the left are squares, and they are both nonnegative. Therefore, the only way their sum can be zero is if both terms are zero. So I can set each term equal to zero and solve for $x$ and $y$. For the first term, I get \[4x^2 = 0 \implies x = 0\] For the second term, I get \[9(y+5)^2 = 0 \implies y+5 = 0 \implies y = -5\] So the only point on the graph is $(0,-5)$. Therefore, the largest possible $y$-coordinate is $-5$. 
The equation \[\frac{x^2}{36} + \frac{(y+5)^2}{16} = 0\]describes a degenerate ellipse, because the right-hand side is $0$ instead of $1$ (as in the standard form for an ellipse). Of all the points on the graph of this equation, what is the largest possible $y$-coordinate?