Function

Let $P(x,y): f(x)f(x+y) = f(2x+y) - xf(x+y)+x$

$\begin{array} {lrll} P(x,x): & f(x)f(2x) & = f(3x) - xf(2x)+x &...(1) \\ P(2x,-x): & f(2x)f(x) & = f(3x) - 2xf(x)+2x &...(2) \\ (2)-(1): & 0 & = 0 - 2xf(x) + xf(2x) + x \\ & \implies f(2x) & = 2 f(x) - 1 & ...(3) \\ & f(0) & = 2 f(0) - 1 \quad \quad \small \color{#3D99F6} \text{Putting }x = 0 \\ & \implies f(0) & = 1 \end{array}$

$\begin{aligned} P(x,0): \qquad f(x)f(x) & = f(2x) - xf(x) + x \\ f^2 (x) & = 2f(x) - 1 - xf(x) + x \\ f^2 (x) - (2-x) f(x) + 1 - x & = 0 \quad \quad \small \color{#3D99F6} \text{A quadratic equation of }f(x) \\ (f(x)-1)(f(x)-1+x) & = 0 \quad \quad \small \color{#3D99F6} \text{As }f(a) \ne f(b) \text{ if }a \ne b \\ \implies f(x) & = 1-x \end{aligned}$

Therefore, $f(150) = 1-150 = \boxed{-149}$

$f(x)=1-x$