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.
