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Given the equation $x = e^{x/y}$, we take the natural logarithm (ln) of both sides to simplify the exponential term.
Using the property $\ln(e^a) = a$, we get: $\ln(x) = \ln(e^{x/y}) = \frac{x}{y}$
From the equation $\ln(x) = \frac{x}{y}$, we solve for $y$: $y = \frac{x}{\ln(x)}$
Now, we differentiate $y$ with respect to $x$ using the quotient rule, which states that if $y = \frac{u}{v}$, then $\frac{dy}{dx} = \frac{v\frac{du}{dx} - u\frac{dv}{dx}}{v^2}$. Here, $u = x$ and $v = \ln(x)$.
We have $\frac{du}{dx} = 1$ and $\frac{dv}{dx} = \frac{1}{x}$. Applying the quotient rule: $\frac{dy}{dx} = \frac{\ln(x) \cdot 1 - x \cdot \frac{1}{x}}{(\ln(x))^2} = \frac{\ln(x) - 1}{(\ln(x))^2}$
Since $\ln(x)$ is the same as $\log_e(x)$, we can rewrite the derivative as: $\frac{dy}{dx} = \frac{\log x - 1}{(\log x)^2}$
Final Answer: $\frac{dy}{dx}=\frac{log~x-1}{(log~x)^{2}}$
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