It may help this: Add side by side, then you get

\partial(f+g)/\partial y = 0, which means that f+g=function of x only. Then, you can assume that

f(x,y) = f0(x) + h(y)

g(x,y) = g0(x) - h(y)

Substituting in the first equation, we get

d(f0(x))/dx - d(h(y))/dy = 0, since the first term depends only on x and the second only on y, then each is equal to a constant, let's say A, hence

f0(x) = Ax + c1; and h(y) = Ay + c2, where c1 and c2 are constants of integration.

Thanks Kamberaj.

How about a∂f/∂x+b∂g/∂y=0 and c∂f/∂y-d∂g/∂x=0, where a,b,c and d are all constants. I think I need a general solution rather than a specific way to a specific problem. But thanks anyway.

You could do the same trick, only after you take first the derivative w.r.t. x variable in the first equation and w.r.t. y variable in the second equation. Then, divide on the first by b\neq 0 and the second by d\neq 0. Then, after you add side by side you get:

\partial^2 f/\partial x^2 = -(cb/da) \partial^2 f/\partial y^2, which is a well known wave equation, and can be solved using variable separation. Similar trick to get an equation for g function.

Dear Shuchang, the two equations are the Cauchy-Riemann conditions applied to h(x,y) = g(x,y) + i f(x,y) (with i2 = -1), therefore h(x,y) is harmonic. Each complex harmonic function h gives solutions g = Re(h) and f = Im(h) to your pdes.  Gianluca