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.
Question 1 (Girsanov's theorem)
Let $W_t$ be a Brownian motion under the physical measure $\mathbb{P}$. Define
$$L_t := \exp \left\{-\int_{0}^{t} X_s dW_s - \frac{1}{2} \int_{0}^{t} X_s^2ds \right\},$$
and define an equivalent martingale measure $\mathbb{Q}$ by setting
$d\mathbb{Q}/d\mathbb{P} = L_t$, then $B_t = W_t + \int_{0}^{t} X_s ds$ is a standard Brownian motion under $\mathbb{Q}$.
Question 2+3
It follows from Girsanov's theorem: for any measurable subset $A$
$$\mathbb{E}_{\mathbb{Q}}[Z] = \int_{A} Z d\mathbb{Q} = \int_{A} Z L_t d\mathbb{P}.$$
Question 4
It's how $L_t$ is defined, show instead by applying Ito's lemma that $dL_t:= -L_t X_t dW_t$.
Question 5
This follows by Ito's formula. If $f$ is a twice differentiable real valued function then
$$f(W_t) = f(W_0) + \int_{0}^{t} f'(W_s)dW_s + \frac{1}{2} \int_{0}^{t} f''(W_s)ds.$$
Apply Ito's formula to $f(x) = x^2$ and the result follows.
Best Answer
If $X_t=F(W_t)$, one knows that $\mathrm dX_t=F'(W_t)\mathrm dW_t+\frac12F''(W_t)\mathrm dt$. If ever there exists some function $F$ such that $$F(0)=1,\qquad F'(w)=-F(w)^2,\qquad F''(w)=2F(w)^3,$$ the proof is complete. Can you identify such a function $F$? Be aware though that there might be no solution $(X_t)$ defined for every nonnegative $t$.