Answer
$\frac{d^3y}{dx^3} = (\frac{d^3y}{du^3})(\frac{du}{dx})^3+3~\frac{d^2y}{du^2}\cdot \frac{du}{dx}\cdot \frac{d^2u}{dx^2}+\frac{dy}{dx}\cdot \frac{d^3u}{dx^3}$
Work Step by Step
$y = f(u) = f(g(x))$
$\frac{dy}{dx} = f'(g(x))\cdot g'(x)$
$\frac{d^2y}{dx^2} = f''(g(x))\cdot g'(x)\cdot g'(x)+f'(g(x))\cdot g''(x)$
$\frac{d^2y}{dx^2} = f''(g(x))\cdot (g'(x))^2+f'(g(x))\cdot g''(x)$
$\frac{d^3y}{dx^3} = f'''(g(x))\cdot g'(x)\cdot (g'(x))^2+f''(g(x))\cdot 2~g'(x)\cdot g''(x)+f''(g(x))\cdot g'(x)\cdot g''(x)+f'(g(x))\cdot g'''(x)$
$\frac{d^3y}{dx^3} = f'''(g(x))\cdot (g'(x))^3+3~f''(g(x))\cdot g'(x)\cdot g''(x)+f'(g(x))\cdot g'''(x)$
$\frac{d^3y}{dx^3} = f'''(u)\cdot (\frac{du}{dx})^3+3~f''(u)\cdot \frac{du}{dx}\cdot \frac{d^2u}{dx^2}+f'(u)\cdot \frac{d^3u}{dx^3}$
$\frac{d^3y}{dx^3} = (\frac{d^3y}{du^3})(\frac{du}{dx})^3+3~\frac{d^2y}{du^2}\cdot \frac{du}{dx}\cdot \frac{d^2u}{dx^2}+\frac{dy}{dx}\cdot \frac{d^3u}{dx^3}$
