I'll assume that $c,d,e$ depend only on $t$. In integral form,
$$
Y_t=Y_0+\int_0^tc_s\,\mathrm ds+\int_0^td_s\,\mathrm dW_s^1+\int_0^te_s\,\mathrm dW_s^2.
$$
Then, for any $C^2$ function $f$ Itô's formula (in differential form) states
$$
\mathrm d\left(f(Y_t)\right)=f'(Y_t)\,\mathrm dY_t+\frac12f''(Y_t)\,\mathrm d\langle Y\rangle_t.
$$
I think your problem is finding $\mathrm d\langle Y\rangle_t$. To do this, use bilinearity and symmetry of $\langle \cdot,\cdot\rangle$:
\begin{multline*}
\langle Y\rangle_t=\langle \int_0^\cdot c_s\,\mathrm ds\rangle_t+\langle \int_0^\cdot d_s\,\mathrm dW_s^1\rangle_t+\langle \int_0^\cdot e_s\,\mathrm dW_s^2\rangle_t
+2\langle \int_0^\cdot c_s\,\mathrm ds,\int_0^\cdot d_s\,\mathrm dW_s^1\rangle_t\\
+2\langle \int_0^\cdot c_s\,\mathrm ds,\int_0^\cdot e_s\,\mathrm dW_s^2\rangle_t+2\langle\int_0^\cdot e_s\,\mathrm dW_s^2,\int_0^\cdot d_s\,\mathrm dW_s^1\rangle_t.
\end{multline*}
Now, $\langle \int_0^\cdot c_s\,\mathrm ds\rangle_t=0$ and $\langle \int_0^\cdot c_s\,\mathrm ds,\int_0^\cdot \varphi_s\,\mathrm dW_s^i\rangle_t=0$ because $\int_0^\cdot c_s\,\mathrm ds$ has finite variation. Also, $\langle\int_0^\cdot e_s\,\mathrm dW_s^2,\int_0^\cdot d_s\,\mathrm dW_s^1\rangle_t=0$ because the two Brownian motions are independent.
The only terms that remain are
$$
\langle Y\rangle_t=\langle \int_0^\cdot d_s\,\mathrm dW_s^1\rangle_t+\langle \int_0^\cdot e_s\,\mathrm dW_s^2\rangle_t
=\int_0^t d_s^2\,\mathrm ds+\int_0^t e_s^2\,\mathrm ds=\int_0^t \left(d_s^2+e_s^2\right)\,\mathrm ds.
$$
Thus,
$$
\mathrm d\left(f(Y_t)\right)=f'(Y_t)\left(c_t\mathrm dt+d_t\mathrm dW^1_t+e_t\mathrm dW^2_t\right)+\frac12f''(Y_t)\left(d_t^2+e_t^2\right)\,\mathrm dt.
$$
Lastly, if you look around, you'll see that a general form of Itô's formula can be applied directly to $f(Y_t)=f\circ F(\int c_s\,\mathrm ds,\int d_s\,\mathrm dW_s^1,\int e_s\,\mathrm dW_s^2)$, making directly appear the quadratic variations of $W_t^1$ and $W_t^2$. The two approaches coincide.