Answer
$\dfrac{d^2y}{dx^2}=\dfrac{6xy-16}{x^3}$.
Work Step by Step
First Derivative:
$\dfrac{d}{dx}(x^2y)-\dfrac{d}{dx}(4x)=\dfrac{d}{dx}(5)\rightarrow$
$2xy+\dfrac{dy}{dx}(x^2)-4=0\rightarrow$
$\dfrac{dy}{dx}=\dfrac{4-2xy}{x^2}$
Second Derivative:
$\dfrac{d}{dx}(\dfrac{dy}{dx})=\dfrac{d}{dx}(\dfrac{4-2xy}{x^2})\rightarrow$
Using the Quotient Rule:
$\dfrac{d^2y}{dx^2}=\dfrac{(-2y+\dfrac{dy}{dx}(-2x))(x^2)-(2x)(4-2xy)}{x^4}\rightarrow$
$\dfrac{2x^2y-2x^3(\dfrac{4-2xy}{x^2})-8x}{x^4}\rightarrow$
$\dfrac{6xy-16}{x^3}.$
