The 'chain rule'. If u= x- y and v= x+ y then [itex]u_x= 1[/itex], [itex]u_v= -1[/itex], [itex]v_x= 1[/itex] and [itex]v_y= 1[/itex].
So [itex]f_x= f_u(u_x)+ f_v(v_x)= f_u+ f_v[/itex] and then [itex]f_{xx}= (f_u+ f_v)_x= (f_u+ f_v)_u+ (f_u+ f_v)v= f_{uu}+ 2f_{uv}+ f_vv[/itex]
Similarly [itex]f_y= f_u(u_y)+ f_v(v_y)= -f_u+ f_v[/itex] and then [itex]f_{yy}= (-f_u+ f_v)_y= -(-f_u+ f_v)_u+ (-f_u+ f_v)_v= f_{uu}- 2f_{uv}+ f_{vv}[/itex].
So [itex]-f_{xx}- f_{yy}= -(f_{xx}+ f_{yy})= -(2f_{uu}+ 2f_{vv})= -2(f_{uu}+ f_{vv})[/itex].
For x and y independent variables, u and v are independent so the integral of that, with respect to x- y= u is [itex]-2f_u+ \Phi(v)[/itex], where [itex]\Phi(v)[/itex] is an arbitrary function of v= x+ y.
(Did your problem really have "-" on both derivatives? With the derivative on only on variable, [itex]u_{xx}- u_{yy}[/itex], this is related to the "wave equation" and subtracting rather than adding gives [itex]4f_{uv}[/itex] and integration with respect to x- y= u gives [itex]4f_v[/itex], or four times the derivative with respect to x+y.)