How Do You Isolate dy/dx in Implicit Differentiation?

[Spiralshell]

Homework Statement

Use implicit differentiation to find an equation of the tangent line to the curve at the given point.

x^2+y^2=(2x^2+2y^2-x)^2 @ (0,(1/2))

Homework Equations

The Attempt at a Solution

So, I don't have a problem differentiating this.

Which I get 2x+2y(dy/dx)=2(2x^2+2y^2-x)(4x+4y(dy/dx)-1)

I am having trouble getting (dy/dx) alone. I was wondering if I could plug in the coordinates and treat (dy/dx) like a variable (e.g. being added and subtracted to other dy/dx like 1(dy/dx)-2(dy/dx)=-1?)

So, could I just plug in (0,(1/2))? Also, I have a test tomorrow and I was wondering in your expert opinions (I know you don't know my professor), but do you think it would be possible that he wants me to simplify this on a test instead (I mean is that more shall I say proper)?

If that is the case when I have 2 large polynomials is it true that I can do something like:
using (2x^2+2y^2-x)(4x+4y(dy/dx)-1) take out 4y(dy/dx) and multiply it by the other polynomial? by that I mean

(2x^2+2y^2-x)(4x+4y(dy/dx)-1) becomes (2x^2+2y^2-x)(4x-1)+4y(dy/dx)(2x^2+2y^2-x)

or like (2x^2+2y^2-x)(4x+4y(dy/dx)-1) is equivalent to (2x^2+2y^2)((4x+4y(dy/dx)-1)-x(4x+4y(dy/dx)-1)?

 

[LCKurtz]

Homework Statement

Use implicit differentiation to find an equation of the tangent line to the curve at the given point.

x^2+y^2=(2x^2+2y^2-x)^2 @ (0,(1/2))

Homework Equations

The Attempt at a Solution

So, I don't have a problem differentiating this.

Which I get 2x+2y(dy/dx)=2(2x^2+2y^2-x)(4x+4y(dy/dx)-1)

I am having trouble getting (dy/dx) alone.

You should probably bone up on your algebra skills then, eh?

I was wondering if I could plug in the coordinates and treat (dy/dx) like a variable (e.g. being added and subtracted to other dy/dx like 1(dy/dx)-2(dy/dx)=-1?)

So, could I just plug in (0,(1/2))?

Yes, that is a very good idea to simplify it quickly, especially when time is at a premium.