Implicit Differentiation of $x^2y = 1$ I have a slight problem with this: Given the equation 
$$ x^2y = 1$$
and asked to find $y''$, I attempted to apply implicit differentiation by differentiation w.r.t. $y$.
$$2xy'y + x^2y' = 0$$
However, it does not seem to be right

UPDATE Solved:
  Differentiate w.r.t. $x$
  $$2xy + x^2y' = 0$$$$
y' = \frac{-2xy}{x^2} = \frac{-2y}{x}
$$

Subsequently,

$$y'' =\frac{x(-2y') - (-2y)(1)}{x^2} = \frac{-2xy' + 2y}{x^2} = \frac{6y}{x^2}$$

 A: We are given that $x^2y = 1 \,\,\,\,\, (\spadesuit)$.
Hence, we have $$\dfrac{d(x^2y)}{dx} = 0 \implies \dfrac{d(x^2)}{dx}y + x^2 \dfrac{dy}{dx} = 0\implies 2xy + x^2 \dfrac{dy}{dx} = 0 \implies 2y + x\dfrac{dy}{dx}=0 \,\,\, (\star)$$
Now differentiate again to get
$$\dfrac{d}{dx} \left(2y + x \dfrac{dy}{dx}\right) = 0 \implies 2 \dfrac{dy}{dx} + \dfrac{d}{dx} \left(x \dfrac{dy}{dx}\right) = 0\\ \implies 2 \dfrac{dy}{dx} + \dfrac{dx}{dx} \dfrac{dy}{dx} + x \dfrac{d^2y}{dx^2} = 0 \implies 3 \dfrac{dy}{dx} + x \dfrac{d^2y}{dx^2} = 0 \,\,\,\, (\dagger)$$
From $(\star)$, we have $\dfrac{dy}{dx} = - \dfrac{2y}x$. Plugging this in $(\dagger)$, we get that
$$x \dfrac{d^2y}{dx^2} + 3 \times \left(- \dfrac{2y}x\right) = 0 \implies \dfrac{d^2y}{dx^2} = \dfrac{6y}{x^2}$$ From $(\spadesuit)$, we have $y = \dfrac1{x^2}$ and hence
$$\dfrac{d^2y}{dx^2} = \dfrac6{x^4}$$
You could do a direct differentiation by noticing that $y = \dfrac1{x^2}$. This implies $$\dfrac{dy}{dx} = -\dfrac2{x^3} \implies \dfrac{d^2y}{dx^2} = \dfrac6{x^4}$$
