Your friends solution is correct. If $(X_t)_{t \geq 0}$ is a one-dimensional Itô process, then Itô's formula states $$ df(t,X_t)= \partial_x f(t,X_t) \, dX_t + \left(\frac{1}{2} \partial_x^2 f(t,X_t) \right) d\langle X \rangle_t + \partial_t f(t,X_t) \, dt. \tag{1}$$ Your friend used this identity for $f(t,x) := x e^{-ct}$.
Your attempt:
$\frac{\partial}{\partial W_t} \left( X_0 e^{ct} + \sigma e^{ct} \int_0^t e^{-cs} \, dW_s \right)$
It this how you are taught to write down Itô's formula? In my oppinion, that's not a good way to write it this way. The problem is that you cannot apply Itô's formula this way. Itô's formula gives you the differential for $f(t,W_t)$ for (nice) functions $f$. But here, you want to calculate the differential of the expression
$$\int_0^t e^{-cs} \, dW_s,$$
i.e. we need a function $f$ such that
$$f(t,W_t) \stackrel{!??!}{=} \int_0^t e^{-cs} \, dW_s.$$
... tell me: How do you choose $f$? Before you have not chosen such a function $f$, you cannot apply Itô's formula this way. What you are doing is treating it as a constant and that's simply not correct.
In order to solve this SDE (or check that the given process is a solution to the SDE) you really have to use Itô's formula for Itô proceses, i.e. $(1)$.
Remark The solution your friend suggested applies Itô's formula to the process
$$e^{-ct} X_t \tag{1}$$
and, at the first glance, it is not obvious how to come up with this particular process. The idea is the following: Instead of considering the SDE
$$dX_t = c X_t \, dt + \sigma \, dW_t$$
we consider the corresponding ODE
$$dx_t = cx_t \, dt$$
(i.e. we just we leave away the stochastic part). It is well-known that the unique solution to this ordinary differential equation is given by
$$x_t = C e^{ct}$$
where $C \in \mathbb{R}$. So far, $C$ is some "deterministic" constant. Now, however, we return to our stochastic setting and allow $C$ to depend on $\omega$ (this is the counterpart of the variation of constants-approach for SDEs). So, by the previous identity, our new auxilary process $C$ is given by
$$C_t = e^{-ct} X_t$$
... and this is exactly the process from $(1)$.
There are a lot of examples where this approach [i.e. first solve the corresponding ODE and then make a "stochastic" variation of constants] works, ee e.g. this question. However, I don't know any statements for which types of SDEs this approach works and for which it doesn't.
Let us take a long route for practice's sake. As we have $u(x,t)=\ln(x^2)$ we can use Ito directly:
$$\frac{\partial u}{\partial t}=0 \ \ \ \ \ \ \ \ \frac{\partial u}{\partial x}=2x\frac{1}{x^2}=\frac{2}{x} \ \ \ \ \ \ \ \ \frac{\partial^2 u}{\partial x^2}=-\frac{2}{x^2}$$
So
$$\begin{aligned}du&=\frac{\partial u}{\partial x}(X_t,t)dX_t+\frac{1}{2}\sigma^2X^2_t\frac{\partial^2 u}{\partial x^2}(X_t,t)dt=\\
&=\frac{2}{X_t}\bigg(\mu X_tdt+\sigma X_tdW_t\bigg)+\frac{1}{2}\sigma^2(-2)dt=\\
&=(2\mu-\sigma^2) dt+2\sigma dW_t \end{aligned}$$
therefore
$$\begin{aligned}u(X_T,T)-u(X_t,t)&=(2\mu-\sigma^2)(T-t)+2\sigma(W_t-W_t)\\
\ln(X_T^2)&=\ln(X_t^2)+(2\mu-\sigma^2)(T-t)+2\sigma(W_t-W_t)\\
\\
\ln(X_T^2)|\mathcal{F}_t&\sim \mathcal{N}(\ln(X_t^2)+(2\mu-\sigma^2)(T-t),\,4\sigma^2(T-t))
\end{aligned}$$
and finally
$$E[\ln(X_T^2)|\mathcal{F}_t]=\ln(X_t^2)+(2\mu-\sigma^2)(T-t)$$
$$\implies u(x,t)=\ln(x^2)+(2\mu-\sigma^2)(T-t)$$
Notice that
$$\ln((X_T/X_t)^2)=2\ln(X_T/X_t)|\mathcal{F_t}\sim \mathcal{N}((2\mu-\sigma^2)(T-t),4\sigma^2(T-t))$$
So $E[\ln((X_T/X_t)^2)|\mathcal{F_t}]=(2\mu-\sigma^2)(T-t)$ and the result you wanted is proved.
Best Answer
This is a quite classical topic, subject of several graduate courses in a lot of universities. A bunch of applications are in Monte Carlo simulation is solutions of PDE in high dimension (where the finite differences methods become inefficient), in particular in Finance.
For advanced extensions to finance, I can recommend you a book from Pierre Henry Labordere: Nonlinear Option Pricing.