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.