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