Exact Differential Equation - Could someone check if I did this right? 
$$ 
\begin{cases}
 \text{ b) }  \cos(x)+ye^{xy}+xe^{xy}\frac {dy}{dx}=0 \\
 \text { c) } e^x+e^y\frac {dy}{dx}=0 \\
\end{cases}
$$

I think I did the problem correctly, but I'm not sure if I wrote the answer in correct form, could someone check?

 A: Looks good to me; $f(x,y)=C$ is the standard way to write solutions to exact differential equations. You could always try to solve explicitly for $y$, and both of those answers are nice enough that you could do that here if you wanted to, but that's usually not necessary.
A: $$ \text{ b) }  \cos(x)+ye^{xy}+xe^{xy}\frac {dy}{dx}=0 $$
I got for this one : $f(x,y)=\sin(x)+e^{xy}=K$
$$ \text { c) } e^x+e^y\frac {dy}{dx}=0 $$
And here $f(x,y)=e^x+e^y=K$
So you did very well for  both equations.
A: $$\cos(x)+ye^{xy}+xe^{xy}\frac {dy}{dx}=0$$
$$\cos(x)+e^{xy}\left(y+x\frac {dy}{dx}\right)=0$$
$\frac{d}{dx}e^{xy}=e^{xy}\left(y+x\frac {dy}{dx}\right)$
$$\cos(x)+\frac{d}{dx}e^{xy}=0$$
The integration leads to 
$$\sin(x)+e^{xy}=C$$
So, your solution is correct. But you can express it explicitly :
$e^{xy}=C-\sin(x)\quad$ so $\quad C-\sin(x)>0 \quad\implies\quad xy=\ln(C-\sin(x))$
$$y=\frac{\ln(C-\sin(x))}{x}$$
$$ $$
Second ODE :
$$e^x+e^y\frac {dy}{dx}=0$$
$\frac {d}{dx}e^y=e^y\frac {dy}{dx}$
$$e^x+\frac {d}{dx}e^y=0$$
The integration leads to 
$$e^x+e^y=C$$
So, your solution is correct. But you can express it explicitly :
$e^{y}=C-e^x\quad$ so $\quad C-e^x>0 \quad\implies\quad $
$$y=\ln(C-e^x)$$
