Is it wrong to derive Log mean temperature difference (LMTD) of a Heat Exchanger, this way?

This is not analyzed correctly. $T_{s1}$ and $T_{s2}$ are not constant, and they need to be eliminated from the analysis all-together. If $dot{Q}$ is the rate of heat flow per unit length along the exchanger, then $$dot{Q}=rho_ih_i(T_h-T_{s1})=rho_oh_o(T_{s2}-T_c)=frac{(rho_i+rho_o)}{2}frac{k_W}{delta}(T_{s1}-T_{s2})$$where $delta$ is the wall thickness and $k_w$ is its thermal conductivity. If we eliminate $T_{s1}$ and $T_{s2}$ from these equations, we obtain: $$dot{Q}=Urho (T_h-T_c)$$Where $$frac{1}{Urho}=frac{1}{h_1rho_i}+frac{1}{frac{(rho_i+rho_o)}{2}frac{k_W}{delta}}+frac{1}{rho_oh_o}$$ Based on this, the heat balance equations for the heat exchanger become $$dot{m}_hC_hfrac{dT_h}{dx}=-Urho(T_h-T_c)$$and$$dot{m}_cC_cfrac{dT_c}{dx}=+Urho(T_h-T_c)$$ I leave it to you to complete the analysis and show that the overall heat load Q is given by:$$Q=(Urho x_f)Delta T_{LM}$$