An iso curve of a function $f(x,y)$ is the path satisfying $f(x,y) = > c$, where $c$ is a given value. Consider a function $f(x, y)$, which is defined for $x\in(-\infty,\infty)$ and $y\in(0,\infty)$. The partial derivatives of $f(x,y)$ are given by
$\frac{\partial f(x,y)}{\partial x}=f_x(x,y)=4e^{4x}y^2$ and $\frac{\partial f(x,y)}{\partial y}=f_y(x,y)=2e^{4x}y$
We assume that the implicit function $y = g(x)$ defined by the isoquant (that is, the iso curve) exists, and that the derivative of $g(x)$ exists. Find the analytical form of $g(x)$ that passes through the point $(0,2)$.
I've been trying to solve this for two hours - I think it's simple but I just can't figure it out. So far, I have used implicit differention on $y=g(x)$ which is has given me the derivative $-2y$. I don't know what to do with the value though. I've tried to integrate the partial derivative to get $e^{4x}y^2+k$ to get some sort of meaning but i'm not sure what to do.
You have obtained $f(x,y)=y^2e^{4x}+k$. The general isoquant is $f(x,y)=c\implies y^2e^{4x}=c-k=m\in\Bbb R$. You want this curve to pass through $(0,2)$ i.e. $m=4$. So the required curve is $y^2e^{4x}=4$ and you can easily convert it to $y=g(x)$ form giving $y=2e^{-2x}$ (we rejected the negative root because given $y>0$).