Functional Equation: When $f(x+y)=f(x)+f(y)-(xy-1)^2$
$$f(x+y)=f(x)+f(y)-(xy-1)^2\\ f(2)=2f(1)\\ f(3)=f(2)+f(1)-1=3f(1)-1\\ f(2)+f(2)-9=f(4)=f(1)+f(3)-4\\ 4f(1)-9=4f(1)-5$$
First put $x=y=0$ into the functional equation to obtain that
$$f(0)=1 \tag{1}$$
Then suppose that $y \ne 0$ so dividing the functional equation by $y$ we can write
$$\frac{f(x+y)-f(x)}{y}=\frac{f(y)-1}{y}-\frac{(xy-1)^2-1}{y} \tag{2}$$
$$\frac{f(x+y)-f(x)}{y}=\frac{f(y)-f(0)}{y-0}-\frac{x^2y^2-2xy}{y} \tag{3}$$
Next, take the limit when $y \to 0$ to obtain
$$f'(x)=f'(0)+2x \tag{4}$$
Where we have assumed that $f(x)$ is differentiable over $\mathbb{R}$. Next, integrating $(4)$ we get
$$f(x)-f(0)=f'(0)x+x^2 \tag{5}$$
Or equivalently
$$f(x)=x^2+f'(0)x+1 \tag{6}$$
But we can observe that $(6)$ will never satisfy the functional equation since $f(x)$ is a second degree polynomial and there are fourth degree terms like $x^2y^2$ in the functional equation . So the only thing that we are left with is that
There is no differentiable $f(x)$ which satisfies the functional equation!