Finding the Maximum value. 
Maximize $xy^2$ on the ellipse $b^2x^2 +a^2y^2= a^2b^2$

The steps I tried to solve:
$$\nabla f = (y^2,2yx)\lambda\qquad g = (2xb^2,2y^2a^2)\lambda$$
$$y^2= 2xb^2\lambda$$
$$2yx= 2y^2a^2\lambda$$

$$
\left.
\begin{array}{l}
\text{}&y^2= 2xb^2\lambda\\
\text{}&
\end{array}
\right\}
*a^2y
$$    
$$
\left.
\begin{array}{l}
\text{}&2yx= 2y^2a^2\lambda\\
\text{}&
\end{array}
\right\}
*b^2x
$$ 

$$y^3a^2= 2yxa^2b^2 \lambda$$      
$$2yx^2b^2 = 2y^2a^2b^2x\lambda$$
$\color{maroon}{\mathbf{Equalize}}$

$$2a^2b^2xy = 2a^2b^2xy^2$$
$$y=1$$
My main problem is which equations does one set each equal to. If anyone knows which one equals the other this will lead me to the right place in finding the solution.
 A: just out of curiosity, we know that $x^2/a^2+y^2/b^2=1$, then by AM–GM inequality
$$1=\frac{x^2}{a^2}+\frac{y^2}{2b^2}+\frac{y^2}{2b^2}\geq 3\sqrt[3]{\frac{x^2}{a^2}\cdot\frac{y^2}{2b^2}\cdot\frac{y^2}{2b^2}}$$
the equality holds when 
$$\frac{x^2}{a^2}=\frac{y^2}{2b^2}=\frac{y^2}{2b^2}$$
A: Since you started off using Langrange multipliers, let’s continue down that path. Using the correct value for $g_y$, we have $$\begin{align}
y^2 &= 2b^2\lambda x \\
2xy &= 2a^2\lambda y
\end{align}$$ which upon eliminating $\lambda$ gives $$
a^2y^3 = 2b^2x^2y,
$$ so either $y=0$ or $a^2y^2=2b^2x^2$. Substituting this into the constraint: $$b^2x^2+2b^2x^2=a^2b^2 \\
\text{or} \\
x^2=\frac13a^2.$$ Plugging this back into the constraint and solving for $y$: $$\frac13a^2b^2+a^2y^2=a^2b^2 \\
\text{or} \\
y^2=\frac23b^2.$$ $f$ is an even function of $y$, so it doesn’t matter which root we use to test these points against $f$. $$
f\left(+\sqrt{\frac13}a,\pm\sqrt{\frac23}b\right) = {2\sqrt3 \over 9}ab^2 \\
\text{and} \\
f\left(-\sqrt{\frac13}a,\pm\sqrt{\frac23}b\right) = -{2\sqrt3 \over 9}ab^2
$$ We still have $y=0$ as a candidate, but $f(x,0)=0$, so this doesn’t yield an extremum.
A: You already have the equation of
$$g(x,y)=b^2x^2+a^2y^2=a^2b^2$$
You can then write $x$ as a function of $y$:
$$x=\pm\frac{1}{b}\sqrt{a^2b^2-a^2y^2}=\pm\frac{1}{ab}\sqrt{a^2-y^2}$$
You can then plug in your values that you found for $y$ from solving the second-to-last equation you gave (note that $y=0$ is a valid solution, as Jack Bauer pointed out, because of the function before you cancelled out the $y$s).
A: First, transform your equation:
$$b^2x^2+a^2y^2=a^2b^2$$
$$a^2y^2=a^2b^2-b^2x^2$$
$$y^2=b^2-\frac{b^2}{a^2}x^2$$
$$xy^2=b^2x-\frac{b^2}{a^2}x^3$$
Next, differentiate and solve to find extrema:
$$b^2-3\frac{b^2}{a^2}x^2=0$$
$$b^2=3\frac{b^2}{a^2}x^2$$
$$a^2=3x^2$$
$$x=\pm\sqrt{\frac{1}{3}}a$$
Finally, evaluate $xy^2$ at $x=+\sqrt{\frac{1}{3}}a$:
$$xy^2=b^2x-\frac{b^2}{a^2}x^3=b^2\frac{a}{\sqrt{3}}-\frac{b^2}{a^2}\frac{a^3}{3\sqrt{3}}=\frac{ab^2}{\sqrt{3}}-\frac{ab^2}{3\sqrt{3}}=\frac{2}{3\sqrt{3}}ab^2$$
That's your maximum. The minimum is at $x=-\sqrt{\frac{1}{3}}a$:
$$xy^2=b^2x-\frac{b^2}{a^2}x^3=-b^2\frac{a}{\sqrt{3}}+\frac{b^2}{a^2}\frac{a^3}{3\sqrt{3}}=-\frac{ab^2}{\sqrt{3}}+\frac{ab^2}{3\sqrt{3}}=-\frac{2}{3\sqrt{3}}ab^2$$
Wolfram Alpha seems to agree.
P.S. Let's also solve the original equation for $y$ at $x=\pm\sqrt{\frac{1}{3}}a$:
$$b^2x^2+a^2y^2=a^2b^2$$
$$b^2\frac{a^2}{3}+a^2y^2=a^2b^2$$
$$\frac{b^2}{3}+y^2=b^2$$
$$y=\pm\sqrt{\frac{2}{3}}b$$
