I know I have to use the method of Lagrange multipliers so I want to find $\lambda$ in
$$\langle 2xy,x^2 \rangle=\lambda \langle2x,4y \rangle$$
This gives me $y =\lambda$ but my multivariable calc is very rusty. Do I use $\lambda$ to find $x,y$ to plug into $x^2y$?
On
Yes express $x,y$ in function of $\lambda$, replace in $x^2+2y^2=6$ to find the value(s) of $\lambda$ then calculate $x^2y$.
Don't forget to check $x=0$ also as pointed out by Michael, and to verify that the obtained value(s) is effectively a maximum (and not a minimum for instance).
For a visualisation make $k$ vary on this drawing:
https://www.desmos.com/calculator/fxuqcug9eq
The maximum is greatest value of $k$ such that the blue curve has intersection points with the ellipse (i.e. when it is tangent, thus the Lagrange multipliers method).
As I mentioned in the comments, if this isn't a homework problem for which you'd lose points for not using Lagrange multipliers, it would be better just to use substitution to change it to a single-variable calculus problem.
In particular, note $x$ only appears as $x^2$ and we may solve $x^2 + 2y^2=6$ for $x^2 = 6-2y^2$. We may now maximize $$x^2y = (6-2y^2)y$$ subject to the condition $6-2y^2 = x^2 \geq 0$. That is, maximize $f(y) = 6y-2y^3$ on the interval $[-\sqrt{3}, \sqrt{3}]$.
Setting the derivative equal to zero to find the critical points gives $$0=f'(y)= 6-6y^2 \implies y=\pm 1$$ and then checking these points gives a maximum of $f(1) = 4$.