Let us take $f(x) = Ax + B$ and substitute this into the functional equation to obtain: $A[x + Axy + By] + B = A[Axy + Bx] + B - x + A[y + Ax + B] + B$; or $Ax + A^{2}xy + ABy + B = A^{2}xy + ABx + 2B + - x + Ay + A^{2}x + AB$; or $Ax + ABy = ABx + B + Ay + A^{2}x - x + AB$. Matching the coefficients for $x, y$, & the constant terms yields: $A = AB + A^2 - 1$ (for x's); $AB = A$ (for y's); $0 = B + AB$ (for constants) which are each satisfied for $A = -1, B = 1$. Thus, the solution is $\fbox{f(x) = 1 - x}$ for all real $x$.

@above, i doult your solution is correct you just proved it for a linear function you have to prove it for any function

Let $P(x,y)$ denote the proposition $f(x + yf(x)) = f(xf(y)) - x + f(y + f(x))$. \begin{align*} P(0,0): f(f(0))&=0 \\ P(f(0),0): f(f(0)^2)&=0 \\ P(f(0)^2,f(0)^2): f(0)^2 &=f(0) \implies \boxed{f(0)=0} \text{ or } \boxed{f(0)=1} \\ \end{align*}If $f(0)=0$, $P(0,x): f(x)=0 $, so contradiction. If $f(0)=1$, $P(x,0): f(f(x))=x$ $P(f(0),x): f(f(x))=1-f(x)\implies \boxed{f(x)=1-x}$ Checking, we see that it works.