<s>[INST] 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 $\frac{dy}{dx}$
$$e^{y/x} \cdot \frac{x-y}{x^2} \frac{dy}{dx} = 20 - \frac{dy}{dx}$$
$$\frac{dy}{dx}+\frac{x-y}{x^2} \frac{dy}{dx} = \frac{20}{e^{y/x}}$$
$$\frac{dy}{dx} + \frac{dy}{dx} = \frac{20}{e^{y/x}\cdot\frac{x-y}{x^2}}$$
$$2\frac{dy}{dx} = \frac{20}{e^{y/x}\cdot\frac{x-y}{x^2}}$$
$$\frac{dy}{dx} = \frac{20}{2e^{y/x}\cdot\frac{x-y}{x^2}}$$
 [/INST] $e^{y/x} \cdot \frac{x-y}{x^2} = 20 - \frac{dy}{dx}$. You forgot that there was a $\frac{dy}{dx}$ term on the left which should have prevented you from simply adding those two fractions together unless you had slid it over into the numerator. But I don't even see it there. Compare your solution to mine:
$$
e^{\frac{y}{x}} = 20x-y\\
\frac{d}{dx}\left(e^{\frac{y}{x}}\right) = \frac{d}{dx}(20x-y)\\
e^{\frac{y}{x}}\frac{d}{dx}\left(\frac{y}{x}\right)=20-\frac{dy}{dx}\\
e^{\frac{y}{x}}\left(\frac{d}{dx}\left(\frac{1}{x}\right)y+\frac{1}{x}\frac{dy}{dx}\right)=20-\frac{dy}{dx}\\
e^{\frac{y}{x}}\left(-\frac{1}{x^2}y+\frac{1}{x}\frac{dy}{dx}\right)=20-\frac{dy}{dx}\\
-\frac{1}{x^2}ye^{\frac{y}{x}}+\frac{1}{x}\frac{dy}{dx}e^{\frac{y}{x}}=20-\frac{dy}{dx}\\
\frac{1}{x}\frac{dy}{dx}e^{\frac{y}{x}}+\frac{dy}{dx}=20+\frac{1}{x^2}ye^{\frac{y}{x}}\\
\frac{dy}{dx}\left(\frac{1}{x}e^{\frac{y}{x}}+1\right)=20+\frac{1}{x^2}ye^{\frac{y}{x}}\\
\frac{dy}{dx}=\frac{20+\frac{1}{x^2}ye^{\frac{y}{x}}}{\frac{1}{x}e^{\frac{y}{x}}+1}\\
\frac{dy}{dx}=\frac{20+\frac{1}{x^2}ye^{\frac{y}{x}}}{\frac{1}{x}e^{\frac{y}{x}}+1}\cdot\frac{x^2}{x^2}\\
\frac{dy}{dx}=\frac{20x^2+ye^{\frac{y}{x}}}{xe^{\frac{y}{x}}+x^2}\\
$$
Wolfram Alpha check.
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