Solow model: Partial derivative of k* with respect to g
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.