Using Lagrange Multipliers to Find Extrema of $f$ I have been trying to solve this lagrange problem for an hour with no result. I can't seem to isolate any variable. 
$$f(x,y) = y^2 -4xy +4x^2$$
With constraint:
$$x^2 + y^2 = 1$$
I have found $x$ to = $-\left(\frac{λ-4}{2}\right)^2 + 1$. Using the constraint $x$ is also equal to $\sqrt{1-y^2}$. This still leaves me with two unknowns. I have tried solving for lambda and setting them equal to each other and I end up with $x^2 = y^2-\frac{3xy}{2}$.
There are still two unknown variables here. I am not sure what I should be isolating to get an answer to work with.

You are given $f(x,y)= y^2- 4xy+ 4x^2$ with constraint $g(x, y)= x^2+ y^2= 1$. Then $\nabla f= <-4y+ 8x, 2y- 4x>$ and $\nabla g= <2x, 2y>$.  The basic idea of the Lagrange multiplier method is that those two vectors are in the same direction- one is a multiple of the other: $<-4y+ 8x, 2y- 4x>= \lambda<2x, 2y>$.  Equating the same components, we have the two equations $4x+ 8y= 2\lambda x$ and $2y- 4x= 2\lambda y$.  Together with the constraint $x^2+ y^2= 1$ that gives three equations for the three unknowns x, y, and $\lambda$.
But a value for $\lambda$ is not necessary for this problem so a good way to solve these equations is to first eliminate $\lambda$ by dividing one of the first two equations by the other:  $\frac{4x+ 8y}{2y- 4x}= \frac{2x+ 4y}{y- 2x}=  \frac{x}{y}$.  From that $y(2+ 4y)= x(y- 2x)$ so $2y^2+ 4xy= xy- 2x^2$, $2y^2+ 3xy+ 2x^2= 0$. Completing the square, $2(x^2+ (3/2)xy+ (9/16)y^2-$$(9/16)y^2)+ y^2= 2(x+ (3/4)y)^2- (9/8)y^2= 0$. x+ (3/4)y= (3/4)y$.   
