Solow model: Partial derivative of k* with respect to g Given the solow model
$sf(k)=(n+g+\delta)k$, written as
$$F=sf(k^\star)-(n+g+\delta)k^\star=0$$
I am trying to find the partial derivative of k* with respect to n
Here is what I've tried: I use the implicit function theorem such that I need to calculate $\frac{\partial k^\star}{\partial n}=-\frac{F_n}{F_{k^\star}}$. I've previously found that
$$F_{k^\star} = sf'(k^\star)-(n+g+\delta)$$
Next I want to find $F_{n}$. Taking the partial derivative and finding that $F_n=-k^\star$. However I am in doubt as to whether or not I should use the chainrule, as $k^{\star}$ is dependant on $n,g,s,\delta$. In that case I find that, because
$$\frac{\partial f(k^\star)}{\partial n}=f'(k^\star)\frac{\partial k^\star}{\partial n}$$
Then we have that $F_n=sf'(k^\star)\frac{\partial k^\star}{\partial n}-k^\star$. Thus we have
$$\frac{\partial k^\star}{\partial n}=-\frac{F_n}{F_{k^\star}}=\frac{sf'(k^\star)\frac{\partial k^\star}{\partial n}-k^\star}{sf'(k^\star)-(n+g+\delta)}$$
It seems odd to find the result of $\frac{\partial k^\star}{n}$ includes itself.
When you are using implicit differentiation along a level curve you will treat the variable with respect to which you are differentiating as a single variable, rather than function.  This is because the formula for implicit differentiation along level curve is already based previous derivation where you already solve for $y'$.
For example, for general function $F(x,y)=c$ where $y=f(x)$, we get by implicit differentiation:
$$v(x) = F(x,y) = c \implies v'(x) = F_x' + F_y' y' = 0 $$
Consequently we get the result that:
$$y' = - \frac{F_x'}{F_y'} \tag{1}$$
Hence, if you just apply formula given by 1 you will not use chain rule anymore, since it was already used in derivation of the formula. For example:
$$xy=5 \implies y'= -\frac{y}{x}  $$
so in your case the correct implicit differentiation along a level curve would yield:
$$\frac{\partial k^*}{\partial n}=  -\frac{F_n'}{F_k'} =\frac{k^*}{sf′(k^⋆)−(n+g+\delta)}$$
You could verify it by actually implicitly differentiating the original function which would yield:
$$\frac{\partial k^*}{\partial n}[sf(k^*)=(n+g+\delta)k^*] \implies sf'(k^*)\frac{\partial k^*}{\partial n} = k^* + (n +g + \delta )\frac{\partial k^*}{\partial n}$$
Solving for $\frac{\partial k^*}{\partial n}$ yields:
$$\frac{\partial k^*}{\partial n} =\frac{k^*}{sf′(k^⋆)−(n+g+\delta)}$$
so it clearly works.