If y=y(x) is an implicit function of x such that

$\ln (x+y)=4 x y$

$($ At $x=0, y=1)$

$x+y=e^{4 x y}$

$\Rightarrow 1+\frac{\mathrm{dy}}{\mathrm{dx}}=\mathrm{e}^{4 \mathrm{xy}}\left(4 \mathrm{x} \frac{\mathrm{dy}}{\mathrm{dx}}+4 \mathrm{y}\right)$

At $x=0$

$\frac{\mathrm{d}^{2} \mathrm{y}}{\mathrm{dx}^{2}}=\mathrm{e}^{4 \mathrm{xy}}\left(4 \mathrm{x} \frac{\mathrm{dy}}{\mathrm{dx}}+4 \mathrm{y}\right)^{2}+\mathrm{e}^{4 \mathrm{xy}}\left(4 \mathrm{x} \frac{\mathrm{d}^{2} \mathrm{y}}{\mathrm{dx}^{2}}+4 \mathrm{y}\right)$

At $x=0, \frac{d^{2} y}{d x^{2}}=e^{0}(4)^{2}+e^{0}(24)$

$\Rightarrow \frac{\mathrm{d}^{2} \mathrm{y}}{\mathrm{dx}^{2}}=40$