$$(x,0)\rightarrow f(x)[f(0)+1]=0\implies f(0)=-1$$$$(2x,-2x)\rightarrow f(2x)f(-2x)=1-\alpha 4x^2$$$$(x,x)\rightarrow f(x)^2=\alpha x^2-f(2x)$$$$(-x,-x)\rightarrow f(-x)^2=\alpha x^2-f(-2x)$$Multiplying two last equations $$(1-\alpha x^2)^2=\alpha^2 x^4-\alpha x^2[f(2x)+f(-2x)]+(1-\alpha 4x^2)$$Hence $$x\neq0\implies -2=f(2x)+f(-2x)$$But for $x=0$ it's true, so for all real $x$ we have $$f(x)+f(-x)=-2$$Using this we solve $$(x,-x)\rightarrow f(x)f(-x)=1-\alpha x^2$$Thus for all real $x$ holds $$\alpha x^2=[f(x)+1]^2$$Now some case work using the initial equation will give $$a>0\implies \forall_{x\in R}f(x)=\sqrt{a}x-1$$$$a<0\implies\text{no solution}$$

WolfusA wrote: $$(x,0)\rightarrow f(x)[f(0)+1]=0\implies f(0)=-1$$$$(2x,-2x)\rightarrow f(2x)f(-2x)=1-\alpha 4x^2$$$$(x,x)\rightarrow f(x)^2=\alpha x^2-f(2x)$$$$(-x,-x)\rightarrow f(-x)^2=\alpha x^2-f(-2x)$$Multiplying two last equations $$(1-\alpha x^2)^2=\alpha^2 x^4-\alpha x^2[f(2x)+f(-2x)]+(1-\alpha 4x^2)$$Hence $$x\neq0\implies -2=f(2x)+f(-2x)$$But for $x=0$ it's true, so for all real $x$ we have $$f(x)+f(-x)=-2$$Using this we solve $$(x,-x)\rightarrow f(x)f(-x)=1-\alpha x^2$$Thus for all real $x$ holds $$\alpha x^2=[f(x)+1]^2$$Now some case work using the initial equation will give $$a>0\implies \forall_{x\in R}f(x)=\sqrt{a}x-1$$$$a<0\implies\text{no solution}$$ I think f(x)=±sqrt(a)*x-1 are both correct.