jmc

Setting $x=y=0$ in the given functional equation, we get $$f(0{)}^2=f(0),$$so $f(0)=0$ or $f(0)=1.$

Setting $x=0,$ we get $$f(0)f(y)=f(y).$$If $f(0)=0,$ then $f(y)=0$ for all $y,$ but this function is not injective. Hence, $f(0)=1.$

Setting $y=x,$ we get $$f(x)f(2x)=f(3x)-xf(2x)+x$$for all $x.$

Setting $x=2t$ and $y=-t,$ we get $$f(2t)f(t)=f(3t)-2tf(t)+2t$$for all $t.$ In other words, $$f(2x)f(x)=f(3x)-2xf(x)+2x$$for all $x.$ comparing this to the equation $f(x)f(2x)=f(3x)-xf(2x)+x,$ we can conlucde that $$-xf(2x)+x=-2xf(x)+2x,$$or $xf(2x)=2xf(x)-x$ for all $x.$ Assuming $x$ is nonzero, we can divide both sides by $x,$ to get $f(2x)=2f(x)-1.$ Since this equation holds for $x=0,$ we can say that it holds for all $x.$

Setting $y=0,$ we get $$f(x{)}^2=f(2x)-xf(x)+x$$Substituting $f(2x)=2f(x)-1,$ we get $$f(x{)}^2=2f(x)-1-xf(x)+x,$$so $$f(x{)}^2+(x-2)f(x)-x+1=0.$$This factors as $$(f(x)-1)(f(x)+x-1)=0.$$Hence, $f(x)=1$ or $f(x)=1-x$ for each individual value of $x.$ If $x\ne 0,$ then $f(x)$ cannot be equal to 1, since $f$ is injective, so $f(x)=\boxed{1-x}.$ Note that this formula also holds when $x=0.$