I have to find the local extrmum of the function $$f(x, y) = x^{2}-2y^{2} - \ln (x + y).$$ I found the first derivatives of $$f'_x = 2x-\frac{1}{x+y}$$ $$f'_y = -4y-\frac{1}{x+y}$$ I made the system $f'_x=0$ and $f'_y = 0$. I found points $A (-1, \frac{1}{2})$ and $B (1, -\frac{1}{2}).$
After that, I found second derivatives $f''_x = 2 + \frac{1}{(x+y)^{2}}$, $f''_y = - 4 + \frac{1}{(x+y)^{2}}, $, $f''_{xy} = \frac{1}{(x+y)^{2}}$, $f''_{yx} = \frac{1}{(x+y)^{2}}$. Then I made The Hessian matrix $\begin{bmatrix} 2 + \frac{1}{(x+y)^{2}} & \frac{1}{(x+y)^{2}}\\ \frac{1}{(x+y)^{2}} & -4 + \frac{1}{(x+y)^{2}} \end{bmatrix}$.
The Problem is that the determinant is $\frac{-8x^{4}-32x^{3}y-48x^{2}y^{2}-2x^2-32xy^3-4xy-8y^4-2y^2}{(x+y)^4}$.
How I can find the local extremum then?