Maximum and minimum value of a function. 
Let us consider a function $f(x,y)=4x^2-xy+4y^2+x^3y+xy^3-4$. Then find the maximum and minimum value of $f$.

My attempt. $f_x=0$ implies $8x-y+3x^2y+y^3=0$ and $f_y=0$ implies $-x+8y+x^3+3xy²=0$ and $f_{xy}=3x^2+3y^2-1$. Now $f_x+f_y=0$ implies $(x+y)((x+y)^2+7)=0$ implies $x=-y$ as $x$ and $y$ are reals. Now putting it in $f_x=0$ get the three values of $x$ as $0$, $(-3+3\sqrt{5})/2$ and $(-3-3\sqrt{5})/2$. And then $f_{xy}(0,0)<0$ implies $f$ has maximum at $(0,0)$. But at other two points $f_{xy}$ gives the positive value. So how can I proceed to solve the problem.
Please help me to solve it.

Note that $f(x,y)=(xy+4)(x^2+y^2-1)$ and therefore
$$\lim_{x\to+\infty}f(x,x)=\lim_{x\to+\infty}(x^2+4)(2x^2-1)=+\infty$$
and
$$\lim_{x\to+\infty}f(x,-x)=\lim_{x\to+\infty}(-x^2+4)(2x^2-1)=-\infty.$$
What may we conclude?
P.S. The critical point $(0,0)$ is a local minimum.
