Is the solution of the boundary value problem correct? I have to solve the following boundary value problem:
$$ u_{xx}+u_{yy}=0, 0<x<a, 0<y<b \tag1$$
$$u(0,y)=u(a,y)=0, 0<y<b $$
$$u(x,0)-u_y(x,0)=0, u(x,b)=f(x), 0<x<a$$
With the method of separation of variables I have found the following:
The solution is of the form $ u(x,y)=X(x)Y(y)$
$$(1) \Rightarrow X''Y+XY''=0 \Rightarrow \frac{X''}{X}+\frac{Y''}{Y}=0 \Rightarrow \frac{X''}{X}=- \frac{Y''}{Y}=- \lambda$$
So we get the following problems:
$$\left.\begin{matrix}
X''+ \lambda X=0, 0<x<a\\ 
X(0)=X(a)=0
\end{matrix}\right\}\tag{*}$$
$$\left.\begin{matrix}
Y''- \lambda Y=0, 0<y<b\\ 
Y(0)-Y'(0)=0
\end{matrix}\right\}\tag{**}$$
$$u(x,b)=X(x)Y(b)=f(x)$$
$(*) \Rightarrow $ The eigenvalues are $\lambda_n =\left(\frac{n \pi}{a}\right)^2$ and the eigenfunction are $ X_n(x)= \sin\left(\frac{n \pi x}{a}\right)$.
$$(**) \Rightarrow Y(y)=A_n \sinh\left(\frac{n \pi y}{a}\right)+B_n \cosh\left(\frac{n \pi y}{a}\right)$$
$$Y'(y)=\frac{n \pi}{a} A_n \cosh\left(\frac{n \pi y}{a}\right)+\frac{n \pi}{a}B_n \sinh\left(\frac{n \pi y}{a}\right)$$
$$Y(0)-Y'(0)=0 \Rightarrow B_n=\frac{n \pi }{a} A_n$$
So $Y_n(y)=A_n \sinh\left(\frac{n \pi y}{a}\right)+\frac{n \pi }{a} A_n \cosh\left(\frac{n \pi y}{a}\right)$
$$u(x,y)=\sum_{n=1}^{\infty}{X_n(x)Y_n(y)}$$
$$ u(x,y)=\sum_{n=1}^{\infty}{A_n\left[ \sinh\left(\frac{n \pi y}{a}\right)+ \frac{n \pi}{a} \cosh\left(\frac{n \pi y }{a}\right)\right] \sin\left(\frac{n \pi x}{a}\right)}$$
$$u(x,b)=f(x) \Rightarrow $$
$$\sum_{n=1}^{\infty}{A_n\left[ \sinh\left(\frac{n \pi b}{a}\right)+ \frac{n \pi}{a} \cosh\left(\frac{n \pi b }{a}\right)\right] \sin\left(\frac{n \pi x}{a}\right)}=f(x) \Rightarrow $$
$$A_n\left[ \sinh\left(\frac{n \pi b}{a}\right)+ \frac{n \pi}{a} \cosh\left(\frac{n \pi b }{a}\right)\right] =\frac{2}{a} \int_0^a{f(x) \sin\left(\frac{n \pi x}{a}\right)}dx $$
$$ \Rightarrow A_n=\frac{2 \int_0^a{f(x) \sin\left(\frac{n \pi x}{a}\right)}dx}{a \sinh\left(\frac{n \pi b}{a}\right)+ n \pi \cosh\left(\frac{n \pi b}{a}\right)}$$
Is my solution correct??

Yes, this is the standard process and you carried it out correctly. I like to write the complete answer as 
$$u(x,y)=\sum_{n=1}^{\infty}{b_n\frac{\sinh\left(\frac{n \pi y}{a}\right)+ \frac{n \pi}{a} \cosh\left(\frac{n \pi y }{a}\right)}{\sinh\left(\frac{n \pi b}{a}\right)+ \frac{n \pi}{a} \cosh\left(\frac{n \pi b }{a}\right)} \sin\left(\frac{n \pi x}{a}\right)} \tag{1}$$
where $b_n$ are the Fourier (sine) coefficients of $f$. Then it's easy to check that all required properties hold: $x=0$ gives $0$, $x=a$  also gives $0$, at $y=0$ we get 
$$u(x,y)=\sum_{n=1}^{\infty}{b_n\frac{ \frac{n \pi }{a}}{\sinh\left(\frac{n \pi b}{a}\right)+ \frac{n \pi}{a} \cosh\left(\frac{n \pi b }{a}\right)} \sin\left(\frac{n \pi x}{a}\right)}$$ for both the function and for its $y$-derivative. And when $y=a$, the sum (1) turns into the sine series of $f$, which is what the coefficients $b_n$ are for.
