Finding maxima and minima of $f(x,y) = x^{4} + y^{4} - x^{3}$ Finding maxima and minima of $$f(x,y) = x^{4} + y^{4} - x^{3}$$
I tried solving this question but need help.
In this question, I calculated the partial derivatives of first and second order.
I found
$$f_{x} = 4x^{3} - 3x^{2}\\   f_{y}= 4y^{3} \\ f_{xx} = 12x^{2}-6x \\ f_{yy}= 0$$
On equating $$f_{x} = 0 \hspace{0.5cm}  and\hspace{0.5cm} f_{y} = 0 $$ I get the critical points as (0,0) and (0.75,0) but I am not able to understand whether these points constitute a maxima or minima because $$f_{xx}f_{yy} - f_{xy}f_{yx}=0$$
Can somebody please guide?

$$f(x,y)=x^4+y^4-x^3\\\qquad\quad ~~ ~=x^3(x-1)+y^4$$
Apply $x\to\  +\infty$ and $y=0$, you get $f(x,y)\to +\infty.$
So, global maxima doesn't exist.

Now, we will prove that global minimum exists and is equal to
$$\text{min}[x^4-x^3+y^2]=-\frac{27}{256}.$$
In this case , of course, $y=0$ must be. Then observe that,
$$\begin{align}x^4-x^3+\frac{27}{256}=\frac{1}{256}(256x^4-256x^3+27)=\frac{1}{256}(4 x - 3)^2 (16 x^2 + 8 x + 3)≥0. \end{align}$$
Because,
$\Delta_{\text{half}}=16-16\times 3=-32<0$.
This means, $16x^2+8x+3>0$  for all $x\in\mathbb R.$
Finally, we deduce that
$$\text{min}\left[x^4-x^3+\frac{27}{256}\right]=0$$
at $x=\dfrac 34.$
which follows
$$\text{min}\left[x^4-x^3+y^4\right]=-\frac{27}{256}.$$
