<s>[INST] Implicit Differentiation Quesrion $$x^y = y^x$$ is the equation
My solution is :$$\frac { dy }{ dx } =\left( \ln { y-\frac { y }{ x }  }  \right) /\left( \ln { x-\frac { x }{ y }  }  \right) $$
Just wondering if this is correct!
 [/INST] Your solution does not match mine:
$$
x^y = y^x \implies x\ln y = y\ln x \implies \frac{\ln y}{y} = \frac{\ln x}{x} \\
\frac{\frac1y\cdot y-1\cdot\ln y }{y^2}dy = \frac{\frac1x\cdot x-1\cdot\ln x }{x^2}dx\\
\frac{1-\ln y}{y^2} dy= \frac{1-\ln x}{x^2}dx\\
\frac{dy}{dx} = \frac{1-\ln x}{1-\ln y}\cdot\frac{y^2}{x^2} 
$$
As I said, that does not seem to reduce to your answer.
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