Let $f(x,y)=x^2y-y^2x$. Find extremus points such that $0\leq x,y \leq 1$. Let $f(x,y)=x^2y-y^2x$.
Find extremus points such that $0\leq x,y \leq 1$.
My attempt:
$\nabla f=(2xy-y^2,x^2-2xy)=0.$
Therefore , the critical points are $(0,0)$.
$H_f= -4x^2-4y^2 \implies H_f(x,y)<0 \forall (x,y)\in [0,1]$.
Am I supposed to check out the points $(0,0),(0,1),(1,0),(1,1)$ ?
$f(0,0)=f(1,1)=0$ , $f(1,0)=1, f(0,1)=-1$.
Hence , maximum :$(1,0)$ , minimum:$(0,1)$.
Is my solution correct ?
 A: You need to be careful with the theory. Note that since the function is a continuous function over a compact set, we can use the extreme value theorem for the case of multivariable functions. This states,

Theorem (Existence of minima and maxima). Let $X\subset \mathbf{R}^{N}$ be a compact subset, and let $f: X\to \mathbf{R}$ be a continuous function. Then the absolute minimum and the absolute maximum of $f$ are attained either at a critical point of $f$ or at boundary point of $X$. (i.e., $f$ attains the absolute maximum and absolute minimum value on $X$.)

Now, since $X:=[0,1]\times [0,1]$ is a compact subset of $\mathbf{R}^{2}$ and the function $f(x,y)=x^{2}y-y^{2}x$ is continuous and defined over $A$, then by the extreme value theorem it attains its absolute maximum and minimum over $X$. Notice that indeed, the critical points of $f$ is given by $(0,0)$ soving the non linear system $\nabla f(a,b)=(0,0)$. Then, if we allow ourselves to sketch the domain $X$, then we can see the following:

This is where the problem with your response arises, as you are only considering the corners of the boundary to draw a conclusion and that is not correct. In terms of the extreme value theorem, you must study the behavior of $f$ over the entire boundary, so it is not enough to analyze only the corners, but in terms of the sketch you should examine what happens with $f$ along $A, B, C$, and $D$. For your particular problem, this is not very difficult.
We can summarize the study of $f(x,y)=x^{2}y-y^{2}x$ on the boundary of $X:=[0,1]\times [0,1]$ as follows
\begin{array}{ccccc}
X& \text{value}& f& \max & \min\\
\hline\\
(0,0)&f(0,0)=\,?&-&-&-\\
(1,0)&f(1,0)=\,?&-&-&-\\
(1,1)&f(1,1)=\,?&-&-&-\\
(0,1)&f(0,1)=\,?&-&-&-\\
A&-&f(x,0)&?&?\\
B&-&f(1,y)&\color{blue}{\frac{1}{4}}&\color{blue}{0}\\
C&-&f(x,1)&?&?\\
D&-&f(0,y)&?&?
\end{array}
Now, to complete the information of $(?)$, the problem is reduced to finding the maxima and minima of single-variable functions, and you can use the methods you know from single-variable calculus to do so. For example, to see the behavior of $f$ along $B$ we have.

*

*Over $B$, we have $x=1$ and $0\leqslant y\leqslant 1$, then $$f(1,y)=y-y^{2},\quad 0\leqslant y\leqslant 1.$$
Then, you can show that (or see directly that it is a quadratic function concave downward). $$\max_{y\in [0,1]}f(1,y)=\frac{1}{4},\quad y=\frac{1}{2}; \quad \min_{y\in [0,1]}f(1,y)=0,\quad y\in \{0,1\}$$
Once you have completed all the missing information in the previous table, then the largest maximum among all the values is the absolute maximum value, and the smallest minimum among all the values is the absolute minimum.
I think you can continue from here, and if you want to verify the answer, you can look inside the following box.

 $$\max_{(x,y)\in [0,1]\times [0,1]}(x^{2}y-y^{2}x)=\frac{1}{4},\quad (x,y)=(1,1/2);$$ $$\min_{(x,y)\in [0,1]\times [0,1]}(x^{2}y-y^{2}x)=-\frac{1}{4},\quad (x,y)=(1/2,1).$$

A: No, that is not correct.  You have a couple of arithmetic errors and have missed four critical points.
Extremal points on a bounded region can be in the interior, where the derivative is $0$, or on the boundary.  So, yes, you need to check the corners as well as the sides, the lines $x= 0$ for $0< y<1$, $x= 1$ for $0< y< 1$,  $y=0$ for $0< x< 1$, and $y= 1$ for $0< x< 1$.
Yes, $\nabla f= \left\langle 2xy−y^2,x^2−2xy\right\rangle= \left\langle 0, 0\right\rangle$ at a critical point so $2xy= y^2$ and, if $y$ is not $0, 2x= y$, and $x^2= 2xy$ so, if $x$ is not $0$, $x= 2y= 2(2x)= 4x$. Yes. there are NO critical points in the interior.  On the line $x= 0$, $f(0,y)= 0$.  On the line $x= 1$, $f(1,y)= y- y^2$ and $f'= 1- 2y= 0$ so $y= 1/2$.  $(1, 1/2)$ is a critical point.  On the line $y= 0$, $f(x,0)= 0$.  On the line $y= 1, f(x,1)= x^2- x$ and $f'= 2x- 1= 0$ so $x= 1/2$. $(1/2, 1)$ is a critical point.
In addition to the corners, $(0, 0)$, $(1,0)$, $(0, 1)$, and $(1, 1)$ we need to check $(1/2, 1)$ and $(1, 1/2)$.
$$\begin{split}
f(0,0)&= 0(0)- 0(0)= 0\\
f(1,0)&= 1(0)- 0(1)= 0 \quad (\text{NOT }1)\\
f(0,1)&= 0(1)- 1(0)= 0 \quad (\text{NOT }-1)\\
f(1,1)&= 1(1)- 1(1)= 0\\
f(1,1/2)&= 1(1/2)- (1/4)1= 1/4\\
f(1/2,1)&= (1/4)1- (1/2)1= -1/4\\
\end{split}$$
The minimum is $-1/4$ at $(1/2,1)$ and the maximum is $1/4$ at $(1,1/2)$.
