Given relation

$x^m y^n = (x+y)^{m+n}$

We need to find $\dfrac{dy}{dx}$.

Take logarithm on both sides

$\ln(x^m y^n)=\ln\big((x+y)^{m+n}\big)$

Using log rules:

$m\ln x+n\ln y=(m+n)\ln(x+y)$

Differentiate implicitly with respect to $x$

$m\cdot \frac{1}{x}+n\cdot \frac{1}{y}\frac{dy}{dx}=(m+n)\cdot \frac{1}{x+y}\left(1+\frac{dy}{dx}\right)$

Rearrange

Multiply throughout by $(x+y)$:

$\frac{m(x+y)}{x}+\frac{n(x+y)}{y}\frac{dy}{dx}=(m+n)\left(1+\frac{dy}{dx}\right)$

Expand:

$m+\frac{my}{x}+\left(n+\frac{nx}{y}\right)\frac{dy}{dx}=m+n+(m+n)\frac{dy}{dx}$

Bring terms together:

$\frac{my}{x}-n=\left(m-\frac{nx}{y}\right)\frac{dy}{dx}$

So,

$\frac{dy}{dx}=\frac{\frac{my}{x}-n}{m-\frac{nx}{y}}$

Simplify

Multiply numerator and denominator suitably:

$\frac{dy}{dx}=\frac{\frac{my- nx}{x}}{\frac{my-nx}{y}}=\frac{y}{x}$

Check with options

$\frac{dy}{dx}=\frac{y}{x}$

Hence the correct option is A.