If $x^m y^n = (x + y)^{m+n}$, then $\frac{d^2y}{dx^2}$ is equal to:

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

Taking natural logarithm of both sides:

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

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

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Differentiating both sides with respect to $x$:

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

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Expanding:

$\frac{m}{x} + \frac{n}{y} \cdot \frac{dy}{dx} = \frac{m+n}{x+y} + \frac{m+n}{x+y} \cdot \frac{dy}{dx}$

$\frac{n}{y} \cdot \frac{dy}{dx} - \frac{m+n}{x+y} \cdot \frac{dy}{dx} = \frac{m+n}{x+y} - \frac{m}{x}$

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

$\frac{dy}{dx} \cdot \frac{nx - my}{y(x+y)} = \frac{nx - my}{x(x+y)}$

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

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Differentiating $\frac{dy}{dx} = \frac{y}{x}$ with respect to $x$:

$\frac{d^2y}{dx^2} = \frac{x \cdot \frac{dy}{dx} - y \cdot 1}{x^2}$

$\frac{d^2y}{dx^2} = \frac{x \cdot \frac{y}{x} - y}{x^2}$

$\frac{d^2y}{dx^2} = \frac{y - y}{x^2}$

$\frac{d^2y}{dx^2} = \frac{0}{x^2}$

$\frac{d^2y}{dx^2} = 0$

Therefore, $\frac{d^2y}{dx^2} = 0$