1. Given implicit relation

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

We need to find $\frac{d^2y}{dx^2}$ at $x=0$.

2. First find the corresponding value of $y$ at $x=0$

Put $x=0$ in the equation:

$\ln(0+y)=4(0)y=0$

So,

$\ln y=0 \implies y=1$

Thus, at $x=0$, we have the point:

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

3. Differentiate once implicitly

Given

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

Differentiate both sides w.r.t. $x$:

$\frac{1+y'}{x+y}=4(xy'+y)$

where $y'=\frac{dy}{dx}$.

Now substitute $(x,y)=(0,1)$:

$\frac{1+y'}{1}=4(0\cdot y'+1)=4$

So,

$1+y'=4 \implies y'=3$

Hence,

$\left.\frac{dy}{dx}\right|_{x=0}=3$

4. Differentiate again

From

$\frac{1+y'}{x+y}=4(xy'+y)$

Differentiate both sides w.r.t. $x$.

For the left side, using quotient rule:

Let

$u=1+y', \quad v=x+y$

Then

$u'=y'', \quad v'=1+y'$

So,

$\frac{d}{dx}\left(\frac{1+y'}{x+y}\right)=\frac{y''(x+y)-(1+y')^2}{(x+y)^2}$

For the right side:

$\frac{d}{dx}[4(xy'+y)] =4(xy''+y'+y')=4(xy''+2y')$

Thus,

$\frac{y''(x+y)-(1+y')^2}{(x+y)^2}=4(xy''+2y')$

5. Substitute $x=0$, $y=1$, $y'=3$

$\frac{y''(1)-(1+3)^2}{1^2}=4(0\cdot y''+2\cdot 3)$

$y''-16=24$

$y''=40$

6. Final answer

$\left.\frac{d^2y}{dx^2}\right|_{x=0}=40$