[Math] Finding the x and y values such that the partial derivatives are zero simultaneously (part two)

Question

$f(x,y)= \ln(x^2 + y^2 +1)$

The partial derivatives of $f(x,y)$ being:

$f_x(x,y) = \frac{2x}{x^2+y^2+1}$ and $f_y(x,y)=\frac{2y}{x^2+y^2+1}$

Setting each partial derivative equal to zero, adding the two linear equations two, and solving for $y$, I get:

$y = – x$

In this problem Finding the $x$ and $y$ values such that the partial derivatives are zero simultaneously, there was only one ordered pair. In the problem before us, I solved it the same way I did in the problem given in the link Does the solution above imply that there are an infinite amount of $(x,y)$ pairs, such that there partial derivatives are zero? I ask, because this particular problem is an even number, and the answer key provides no answers for such questions.

Answer 1

Hint: Since the denominators of these partial derivatives are never zero, then the partial derivatives are equal to zero precisely when their numerators are.