Help with implicit differentiation simplifications STEP 1: $$ (x+y)^{3} = x^3 + y^3 $$ 
STEP 2: $$ 3(x+y)^2 (1+ dy/dx) = 3x^2 + 3y^2(dy/dx) $$
STEP 3: $$ 3(x+y)^2 + \frac{dy}{dx}\cdot3(x+y)^2 = 3x^2 + 3y^2(dy/dx)$$
STEP 4: $$\frac{dy}{dx} \cdot 3(x+y)^2 = 3x^2 + 3y^2(dy/dx) - 3(x+y)^2$$
STEP 5: $$\frac{dy}{dx} = (3x^2 + 3y^2(dy/dx))/3(x+y)^2 - 1$$
I do not quite get how to move the dy/dx to one side, and all the others to the 
    other side to solve for what dy/dx is.
It is one of my worst weaknesses when I try to do an implicit differentiation.
    I've searched thorugh wolfram alpha to see how such simplification works,
    but some of the problems seems to only have simple step-by-step process shown,
    not the whole step-by-step solution.

Hint $x^3 + y^3 = (x+y)(x^2-xy+y^2)$
so we have the expression
$$(x+y)^3=(x+y)(x^2-xy+y^2)$$
$$(x+y)[(x+y)^2-(x^2-xy+y^2)]=0$$
$$(x+y)(3xy)=0$$
$$x^2y+y^2x=0$$
which when differentiated becomes
$$x^2\displaystyle\frac{dy}{dx}+y(1)+y^2(1)+x\displaystyle\frac{dy}{dx}=0$$
$$\displaystyle\frac{dy}{dx}(x^2+x)=-(y^2+y)$$
$$\displaystyle\frac{dy}{dx}=-\displaystyle\frac{y^2+y}{x^2+x}$$
