Differentiate $e^{y/x} = 20x-y$ I am trying to use implicit differentiation to differentiate $e^{y/x} = 20x-y$. I get $\frac{20}{2 e^{y/x} \cdot \frac{x-y}{x^2}}$, but according to the math website I'm using, "WebWork", this is wrong. 
I'm not sure how to handle it when an equation has two $\frac{dy}{dx}$ floating around, but I'm not sure this is the problem.
Here are my steps:
$$\frac{d}{dx}(e^{y/x}) = \frac{d}{dx}(20x-y)$$
Factoring both sides independently:
$$\frac{d}{dx}(e^{y/x}) = e^{y/x}\cdot\frac{x-y}{x^2} \frac{dy}{dx}$$
$$\frac{d}{dx}(20x-y)=20-\frac{dy}{dx}$$
Finding $\frac{dy}{dx}$
$$e^{y/x} \cdot \frac{x-y}{x^2} \frac{dy}{dx} = 20 - \frac{dy}{dx}$$
$$\frac{dy}{dx}+\frac{x-y}{x^2} \frac{dy}{dx} = \frac{20}{e^{y/x}}$$
$$\frac{dy}{dx} + \frac{dy}{dx} = \frac{20}{e^{y/x}\cdot\frac{x-y}{x^2}}$$
$$2\frac{dy}{dx} = \frac{20}{e^{y/x}\cdot\frac{x-y}{x^2}}$$
$$\frac{dy}{dx} = \frac{20}{2e^{y/x}\cdot\frac{x-y}{x^2}}$$

You get:
$$(e^{y/x})'_x=(2x-y)'_x \Rightarrow \\
e^{y/x}\cdot (y/x)'_x=2-y'\Rightarrow \\
(2x-y)\cdot \frac{y'x-y}{x^2}=2-y' \Rightarrow \\
(2x^2-xy)y'-(2x-y)y=2x^2-x^2y'\Rightarrow \\
y'=\frac{2x^2+2xy-y^2}{3x^2-xy}.$$
If you are familiar with multivariable calculus, consider: $F(x,y)=e^{y/x}-2x+y=0$. Then:
$$y'=-\frac{F_x}{F_y}=-\frac{e^{y/x}\cdot (-y/x^2)-2}{e^{y/x}\cdot (1/x)+1}=\frac{(2x-y)y+2x^2}{(2x-y)x+x^2}=\frac{2x^2+2xy-y^2}{3x^2-xy}.$$
