For two-variables function, critical points are defined as the points in which the gradient equals zero, just like you had a critical point for the single-variable function f(x) if the derivative f'(x)=0. The matter is that you now can differentiate the function with respect to more than one variable (namely 2, in your case), and so you must define a derivative for each directions.
The gradient is thus defined as the n-dimensional vector (again, in your case n=2), and its coordinates are the derivatives with respect to each variable. So, the gradient of a two-variable function f(x,y) is the vector
(\frac{\partial f}{\partial x}, \frac{\partial f}{\partial y}), where deriving with respect to a variable means to consider the other as a constant.
Let's compute the two derivatives:
\frac{\partial f}{\partial x}= 24 x^2 + 144y
\frac{\partial f}{\partial y}= 144x + 24 y^2
To find the critical points, we must find the values of x and y for which
(\frac{\partial f}{\partial x}, \frac{\partial f}{\partial y})=(0,0) holds.
In other words, we must solve
24 x^2 + 144y=0
24 y^2 + 144x=0
Simplifying both expression, we have
x^2 + 6y=0
y^2 + 6x=0
An obvious solution is (x,y)=(0,0), which is thus our first critical point, while the other solution are (x,y)=(-6,-6), which is the second and last critical point.
