If ${e^y}(x + 1) = 1,$ then show that $\cfrac{{{d^2}y}}{{d{x^2}}} = {\left( {\cfrac{{dy}}{{dx}}} \right)^2}$ .

$x{e^y} + {e^y} = 1$ ...(i)
Differentiating (i) w.r.t. x, we get

$x{e^y}\cfrac{{dy}}{{dx}} + {e^y} + {e^y}\cfrac{{dy}}{{dx}} = 0$ $\Rightarrow$

$\cfrac{{dy}}{{dx}} = \cfrac{{ - 1}}{{x + 1}}$ …(ii)

From (ii), ${\left( {\cfrac{{dy}}{{dx}}} \right)^2} = {\left( {\cfrac{{ - 1}}{{x + 1}}} \right)^2} = \cfrac{1}{{{{(x + 1)}^2}}}$ …(iii)

Differentiating (ii) w.r.t. x, we get
$\cfrac{{{d^2}y}}{{d{x^2}}} = \cfrac{1}{{{{(x + 1)}^2}}}$ ..(iv)

From (iii) and (iv), we get
$\cfrac{{{d^2}y}}{{d{x^2}}} = {\left( {\cfrac{{dy}}{{dx}}} \right)^2}$

Hence proved. .