Put both derivatives equal to zero: then you have a system
$$
3x^2-63+12y=0 \\
3y^2-63+12x=0
$$
Isolate $y$ in the first equation and you get $y=-\frac{1}{4}x^2+\frac{21}{4}$. Substitute it in the second equation to get $3(-\frac{1}{4}x^2+\frac{21}{4})^2-63+12x=0$ or equivalently $$\frac{3}{16}x^4-\frac{63}{8}x^2+12x-63+3\cdot \frac{441}{16}=0\\
\frac{3}{16}(x^4-42x^2+48x-336+441)=0\\
\frac{3}{16}(x^4-42x^2+64x+105)=0\\
\frac{3}{16}(x-5)(x-3)(x+1)(x+7)=0$$
Now you can find the corresponding $y$'s and the solutions of the systems are $(-7,-7),(-1,5),(3,3),(5,-1)$
You just need to classify them now
My first advice is to learn latex my friend, it's very helpfull!
Lets consider $$f(x, y) = x^3 - y^3 - 12xy -63x + 63y$$
Taking both partial derivate of $x$ and $y$ and making them equal $0$,
$$\frac{\partial}{\partial x} f(x, y) = 3x^2-12y-63 = 0, \quad (1)$$ and $$\frac{\partial}{\partial y} f(x, y) = -3y^2 -12x +63x = 0, \quad (2)$$
Clearing $y$ from equation $(1)$ and substituting in equation $(2)$, $$y=\frac{3x^2-63}{12} \Longrightarrow -3\Big(\frac{3x^2-63}{12}\Big)^2-12x+63=-\frac{3x^4}{16}+\frac{63x^2}{8}-12x+\Big(63-\frac{1323}{16}\Big)=0$$ which is a fourth degree polynomial, then $$x\in\{-7, -1, 3, 5\}$$ Using this set of points for $x$ on equation $(1)$ (or $(2)$, it's the same), were have the following set of stationary points, $$A = \{(-7, 7), (-1, -5), (3,-3), (5, 1)\}$$
Let's consider the following Theorem.
Let $$D=\frac{\partial^2}{\partial x^2} f(a, b) \cdot \frac{\partial^2}{\partial y^2} f(a, b) -\Big(\frac{\partial^2}{\partial x \partial y} f(a, b)\Big)^2, \quad (a, b) \in A$$ then,
$1)$ If $D>0$ and $\frac{\partial^2}{\partial x^2} f(a, b) > 0$, then $f$ has a minimum at $(a, b)$.
$2)$ If $D>0$ and $\frac{\partial^2}{\partial x^2} f(a, b) < 0$, then $f$ has a maximum at $(a, b)$.
$3)$ If $D<0$, then $f$ has a saddle point at $(a, b)$.
$4)$ If $D=0$, then we have no conclusion for $f$ at $(a, b)$.
Using this Theorem with the points of the set $A$, we conclude that, $(-7, 7)$ is maximum of $f$, $(3, -3)$ is minimum of $f$ and both $(-1, -5)$ and $(5, 1)$ are saddle points of $f$.
Best Answer
By equating we obtain
$$3x^2-3y^2=12y-12x\iff3(x-y)(x+y)=12(y-x)\iff y=x \quad \lor \quad x+y=4$$
and thus
$$3x^2-63+12x=0\implies x=-7,3 \implies (x,y)=(-7,-7) \quad (x,y)=(3,3)$$
$$3x^2-63+12(4-x)=0\implies x=-1,5 \implies (x,y)=(-1,5) \quad (x,y)=(5,-1)$$