I think you meant to say
[tex]\ln(x+y) = 4xy[/tex]
and not "4 times y" on the right side (which would lead to a complex value for y when x = 0). Note that when x = 0, the equation reduces to ln(y) = 0, so that y = 1.
Implicitly differentiating both sides with respect to x, taking y = y(x), and solving for dy/dx gives
[tex]\dfrac{1+\frac{dy}{dx}}{x+y} = 4y + 4x\dfrac{dy}{dx}[/tex]
[tex]\implies \dfrac{dy}{dx} = \dfrac{4xy+4y^2-1}{1-4x^2-4xy}[/tex]
Note that when x = 0 and y = 1, we have dy/dx = 3.
Differentiate both sides again with respect to x :
[tex]\dfrac{d^2y}{dx^2} = \dfrac{(1-4x^2-4xy)\left(4y+4x\frac{dy}{dx}+8y\frac{dy}{dx}\right)-(4xy+4y^2-1)\left(-8x-4y-4x\frac{dy}{dx}\right)}{(1-4x^2-4xy)^2}[/tex]
No need to simplify; just plug in x = 0, y = 1, and dy/dx = 3 to get
[tex]\dfrac{d^2y}{dx^2} \bigg|_{x=0} = \boxed{40}[/tex]
