Using implicit differentiation with a fraction How do I solve this? What steps?  I have been beating my head into the wall all evening. 
$$ x^2 + y^2 = \frac{x}{y} + 4 $$
 A: We have:
$$ \frac{d(x^2)}{dx} + \frac{d(y^2)}{dy}\cdot \frac{dy}{dx} = \frac{d( \frac{x}{y})}{dx} + \frac{d(4)}{dx}$$
$$2x+2y\frac{dy}{dx} = \frac{y-x \frac{dy}{dx} }{ y^2} +0$$
Can you finish?

Okay, to finish it up:
$$2y\frac{dy}{dx} - \frac{y-x \frac{dy}{dx} }{ y^2} = -2x  $$
Multiplying both sides by $y^2$
$$2y^3\frac{dy}{dx}-(y-x\frac{dy}{dx})=-2xy^2$$
$$2y^3\frac{dy}{dx}+x\frac{dy}{dx}=-2xy^2 +y$$
$$\frac{dy}{dx} \cdot (2y^3+x) =-2xy^2 +y$$
$$\frac{dy}{dx} = \frac{-2xy^2 +y}{2y^3+x}$$
A: We can multiply both sides of the equation
$$x^2 + y^2 = \frac{x}{y} + 4 \tag{1}$$
by $y$ to obtain
$$x^2y + y^3 = x + 4y \tag{2}$$
Differentiating equation 2 implicitly with respect to $x$ yields
\begin{align*}
2xy + x^2y' + 3y^2y' & = 1 + 4y'\\
(x^2 + 3y^2 - 4)y' & = 1 - 2xy\\
y' & = \frac{1 - 2xy}{x^2 + 3y^2 - 4} \tag{3}
\end{align*}
Let's see why this answer is equivalent to that given by @CivilSigma.  If we solve equation 1 for $x/y$, we obtain
$$\frac{x}{y} = x^2 + y^2 - 4$$
Hence,
\begin{align*}
y' & = \frac{1 - 2xy}{x^2 + 3y^2 - 4}\\
   & = \frac{1 - 2xy}{x^2 + y^2 - 4 + 2y^2}\\
   & = \frac{1 - 2xy}{\dfrac{x}{y} + 2y^2}\\
   & = \frac{y - 2xy^2}{x + 2xy^3}
\end{align*}
