# Classifying critical points of Multivariate function

Looking to find critical points and classify them as max/min or saddles for the following multivariate function

$$f(x,y)=x^2y+y^3-48y$$

Computed the partial derivatives with respect to $$x$$ and $$y$$ and equated them to zero and got the following critical points $$(-4\sqrt(3),0), (4\sqrt(3),0),(0,-4),(0,4)$$

Using the formula for second derivative test I found that all critical points above were saddle points except for $$(0,4)$$ which turned out to be a min as $$D(0,4)>0$$ and $$f_{xx}>0$$.

After doing so I found that apparently there is still something wrong with this answer. Can anyone give a hint as to what that is as I have spent a long time at this and still can't find it?

$$\frac{\partial^2f}{\partial x^2}\cdot\frac{\partial^2f}{\partial y^2}-\left(\frac{\partial^2f}{\partial x\partial y}\right)^2=2y\cdot6y-(2x)^2=4(3y^2-x^2),$$ which gives that $$(\pm4\sqrt3,0)$$ are not extreme points.
Now, $$\frac{\partial^2f}{\partial x^2}(0,4)>0,$$ which gives that $$(0,4)$$ is a minimum point and $$\frac{\partial^2f}{\partial x^2}(0,-4)<0,$$ which gives that $$(0,-4)$$ is a maximum point.