Use Lagrange multipliers to find the max and min of the function $f(x,y)=xe^y$ subject to the constraint $x^2+y^2=6$. 
Use Lagrange multipliers to find the max and min of the function $f(x,y)=xe^y$ subject to the constraint $x^2+y^2=6$.

My attempt:
I tried to find both partials and set them equal to $\lambda$ times the partial of the constraint and got the following equations
$$e^y=2x(\lambda)$$
$$x(e^y)=2y(\lambda) $$
I then solved for $x$ and $y$ and got :
 $$x=\frac{e^y}{2\lambda}$$
$$y=\frac{xe^y}{2\lambda}$$
I then plugged those into the original equation and got 
$$\frac{(x^2+1)(e^(2y) )}{4\lambda^2}=6$$
I'm confused where to go from here and any help will be greatly appreciated
 A: Following your calculations:
$$2\lambda=e^y/x$$
$$2\lambda=xe^y/y$$
so
$$e^y/x=xe^y/y \iff ye^y=x^2e^y \iff y=x^2$$ where in the last equivalency we use that $e^y\ne0$.
Now, use this information with your constraint to get a bicuadratic equation on $x$ that you can solve alone.
A: We have,
$$e^y = 2 \lambda x$$
$$x e^y = 2 \lambda y$$
thus,
$$x = \frac{2 \lambda y}{e^y} = \frac{2 \lambda y}{2 \lambda x} = \frac{y}{x}$$
finally,
$$x^2 = y$$
Now we replace into the original (constraint) equation 
$$y^2 + y - 6 = 0$$
You have now the constraint in the form of a quadratic equation, you can easily find the value of $y$ through solving the equation.
A: Without Lagrange Multipliers
Calling
$$
x = r\cos(\theta)\\
y = r\sin(\theta)
$$
we have the equivalent problem
$$
\max\min_{\theta}f(\theta) = \sqrt{6}\cos(\theta) e^{\sqrt 6\sin(\theta)}
$$
and
$$
f'(\theta) = \sqrt{6}e^{\sqrt{6} \sin (\theta )} \left(\sqrt{6} \cos ^2(\theta )-\sin (\theta )\right)\to \sqrt{6} \cos ^2(\theta )-\sin (\theta ) = 0
$$
etc.
NOTE
This result is equivalent to the system
$$
x^2+y^2=6\\
x^2-y = 0
$$
