Let us first recall some definitions and useful formulas for a surface in $\mathbb{R}^3$ given by an immersion $F\colon U \rightarrow \mathbb{R}^3$, where $U \subset \mathbb{R}^2$ is an open set.
Denote $F_x = \frac{\partial F}{\partial{x}}$, $F_{x x} = \frac{\partial{F_x}}{\partial{x}}$, and so on.
The components of the first fundamental form are $I_{x x} = F_x \cdot F_x$, $I_{y y}=F_y \cdot F_y$, $I_{x y} = I_{y x} = F_x \cdot F_y$
For the second fundamental form we may use the expression for the unit normal vector
$$
n = \frac{F_x \times F_y}{|F_x \times F_y|}
$$
so that $II_{x x} = n \cdot F_{x x}$, $II_{y y} = n \cdot F_{y y}$, $II_{x y} = II_{y x} = n \cdot F_{x y}$
The Gaussian curvature $K(x,y)$ has the following expression
$$
K(x,y) = \frac{II_{x x} II_{y y} - II_{x y}^2}{I_{x x} I_{y y} - I_{x y}^2} \tag{1}
$$
and the mean curvature can be computed by
$$
H(x,y) = \frac{I_{x x} II_{y y} - 2 I_{x y} II_{x y} + I_{y y} II_{x x}}{I_{x x} I_{y y } - I_{x y}^2} \tag{2}
$$
In the proposed problem we a given a surface represented by a graph of function $z = f(x) + g(y)$, so our immersion has the following form:
$$
F(x,y) = \begin{pmatrix}
x \\
y \\
f(x) + g(y)
\end{pmatrix}
$$
and we calculate $F_x = \begin{pmatrix} 1 \\ 0 \\ f' \end{pmatrix}$, $F_y = \begin{pmatrix} 0 \\ 1 \\ g' \end{pmatrix}$, $F_{x x} = \begin{pmatrix} 0 \\ 0 \\ f'' \end{pmatrix}$, $F_{y y} = \begin{pmatrix} 0 \\ 0 \\ g'' \end{pmatrix}$, $F_{x y} = \begin{pmatrix} 0 \\ 0 \\ 0 \end{pmatrix}$.
This is enough to find the unit normal
$$
n = \frac{(f', -g', 1)^T}{\sqrt{1 + (f')^2 + (g')^2}}
$$
so we find the components of the second fundamental form $II_{x x} = \frac{f''}{\sqrt{1 + (f')^2 + (g')^2}}$, $II_{y y} = \frac{g''}{\sqrt{1 + (f')^2 + (g')^2}}$, $II_{x y} = 0$
The components of the first fundamental form are, of course, $I_{x x} = 1 + (f')^2$, $I_{y y} = 1 + (g')^2$, and $I_{x y} = f'g'$.
I would let you to finish the job by substituting these quantities into equations (1) and (2).
While it's true that the computation of $K$ is much easier in orthogonal and especially in isothermal coordinates, finding such coordinates requires solving a PDE. I don't think it would be worth the effort in this case. Your coefficients $E,F,G$ are about as simple as one could have, the partials are immediately found and the determinants are not too bad. Everything is quite doable by hand.
With apologies for using a proprietary system (Maple), here is the general Brioschi formula:
with(LinearAlgebra):
E := 1+v^2; F := 2*u*v; G := 1+u^2;
A := Matrix([[-diff(E,v,v)/2+diff(F,u,v)-diff(G,u,u)/2, diff(E,u)/2, diff(F,u)-diff(E,v)/2], [diff(F,v)-diff(G,u)/2, E, F], [diff(G,v)/2, F, G]]);
B := Matrix([[0, diff(E,v)/2, diff(G,u)/2], [diff(E,v)/2, E, F], [diff(G,u)/2, F, G]]);
K := simplify((Determinant(A)-Determinant(B))/(E*G-F^2)^2);
Output: $$K=\frac{u^2+v^2}{(1+u^2+v^2-3u^2v^2)^2}$$
Best Answer
Look here
http://en.wikipedia.org/wiki/Gaussian_curvature
under the subheading "Alternative formulas". If you want a proof, compute the first and second fundamental forms first.