Find the extreme values of the function subject to the given constraint

$\displaystyle f(x,y)=x^{2}y$, $\displaystyle x^{2}+2y^{2}=6$

Ok, so I'm trying to do this using Lagrange multipliers

So, first I need $\displaystyle \nabla f=\lambda \nabla g$, right?

Which is:

$\displaystyle 2xyi+x^{2}j=\lambda (2xi+4yj)$

So:

$\displaystyle 2xy=\lambda 2x$ and $\displaystyle x^{2}=\lambda 4y$

From here, I have no idea what to do. Any help?

Edit:

Ok, I THINK I figured it out. Please someone double check my work

So,

$\displaystyle y=\lambda$ and $\displaystyle x^{2}=4\lambda^{2}$

We plug these into the constraint and get:

$\displaystyle 4\lambda^{2}+2\lambda^{2}=6$

$\displaystyle =>\lambda=\pm 1$

Then we plug $\displaystyle \lambda$ back into $\displaystyle y=\lambda$ and $\displaystyle x=2\lambda$

So we get, $\displaystyle y=\pm 1$ and $\displaystyle x=\pm 2$

Then, plugging these numbers into $\displaystyle f(x,y)=x^{2}y$

$\displaystyle (-2)^{2}(-1)=-4$ and $\displaystyle (2)^{2}(1)=4$

Is this all correct?