The first order Taylor expansion of this system of equations in the picture if pretty straightforward. However I don't get how we get values for the steady state.
Is it correct to say that in the steady state:
(1) $\dot k = 0$, $\dot c = 0$
(2) as well as the partial derivative of both $\dot c$, $\dot k$ with respect to either $c$ or $k$, is zero as well?
Do you have to calculate the SS: $(k^*,c^*)$ based on equation (1) or one of the equations in (2), as all of these would give different results, no?
Because I don't get why the partial derivative of $\dot k$ with respect to $k$ would equal $\rho$? See extra notes to see what this equals to. I would think it would be zero?
If that is the case, I get that in the SS, $k$ is $k^* = (\delta + \rho/\alpha)^{1/(\alpha-1)}$, but I don't get why:
$$
c^* = (k^*)^\alpha - \delta k^*.
$$
Attachment: 
The steady states you determine with (1). The stability of these steady states you examine with the Jacobian that is built from the partial derivatives as indicated in (2).
By context, it seems that of the 4 steady states the one with non-zero $c^*$ is chosen. Then the non-trivial factor in the first equation gives $$ \dot c=0\land c^*\ne 0\implies k^*=\left(\frac{δ+ρ}α\right)^{1/(α−1)} $$ From the non-trival factor in the second equation you get, using the above value for $k^*$, $$ \dot k=0 \implies c^*=(k^∗)^α−δk^∗. $$