Functional Soiree!

Given that

$\begin{aligned} f(x)f(y) + 2 & = f(x) + f(y) + f(xy) & \small \blue{\text{Taking }\frac {\partial}{\partial y} \text{ on both sides}} \\ f(x)f'(y) & = f'(y) + xf'(xy) & \small \blue{\text{Putting }x=0 \text{ and }y=1} \\ f(0)f'(1) & = f'(1) + (0)f'(0) \\ 3f(0) & = 3 \\ \implies f(0) & = \boxed 1 \end{aligned}$

Since the first equation holds for all x and y, it holds for y = 0. This means:

f(x) f(0) + 2 = f(x) + 2 f(0), or

f(0)*(f(x) - 2) = (f(x) - 2) for every value of x.

Unless f(x) is identically 2, division by f(x) - 2 would mean f(0) = 1. However, f(x) can't be identically equal to 2, or else the derivative would be identically zero (constant function), contrary to the fact that f'(1) = 3. Therefore, f(0) = 1.

Differentiating this functional equation with respect to $x$ and $y$ respectively yields:

$f'(x)f(y) = f'(x) + yf'(xy)$ (i)

$f(x)f'(y) = f'(y) + xf'(xy)$ (ii)

or $\frac{f'(x)[f(y)-1]}{y} = f'(xy) = \frac{f'(y)[f(x)-1]}{x};$

or $\frac{f'(x)}{f(x)-1} = \frac{A}{x}$ (for $A \in \mathbb{R});$

or $\ln[f(x)-1] = A\ln(x) + B;$

or $f(x) -1 = e^{B}x^{A}$ ;

or $f'(x) = Ae^{B}x^{A-1}$ .

Plugging in the above two boundary conditions, we obtain :

$2 - 1 = e^{B} \cdot 1^{A}, 3 = Ae^{B} \cdot 1^{A-1} \Rightarrow A =3, B = 0$

or $f(x) = x^3 + 1$ . Thus, $\boxed{f(0)=1}.$

Let $h(x) = f(x)-1.$ Then the first equation becomes $\begin{aligned} f(x)f(y)-f(x)-f(y)+1 &= f(xy)-1 \\ (f(x)-1)(f(y)-1) &= f(xy)-1 \\ h(x)h(y) &= h(xy) \end{aligned}$ This implies that $h(0)h(y) = h(0)$ for all $y.$ So either $h(y) = 1$ for all $y,$ or $h(0) = 0.$ But $h$ is not a constant function, because of the assumption about the derivative, so $h(0) = 0,$ so $f(0) = \fbox{1}.$

Set $y=x-1$ , and taking the derivative with respect to $x$ yields $f'(x)f(x-1)+f(x)f'(x-1)=f'(x)+f'(x-1)+(2x-1)f'(x^2-x).$ Setting $x=1$ gives $3f(0)+2f'(0)=3+f'(0)+f'(0)\implies f(0)=\boxed{1}$

If f’(y)=x then f(x)=y Which implies f(3)=1 From the definition f(3) f(0)+2=f(3)+f(0)+f(0 3) 1*f(0)+2=2f(0)+1 Therefore f(0)=1

plugging in $x=y=0$ gives a quadratic with solutions $f(0) = 1$ or $f(0)=2$ . Which of these is it?

Differentiate w.r.t. x keeping y=0: $d(f(x)f(0)+2)=d(f(x)+df(0)+df(0))$

$f(0)f'(x)dx+0=f'(x)dx+0+0$

plugging in x=1 and dividing by dx gives $3f(0)=3$

so $f(0)=1$

Is my approach correct?

$f'(1)=3 \implies f(3)=1$

Putting $x=0,y=3$ $f(0)+2=f(0)+1+f(0)$ $\implies f(0)=1$

$f(x)f(y) + 2 = f(x) + f(y) + f(xy) \\ f(0)f(0) + 2 = f(0) + f(0) + f(0 \times 0) \\ f(0)^2 - 3f(0) + 2= 0 \\ (f(0) - 1)(f(0) - 2) = 0 \\ \therefore f(0) = 1 or f(0) = 2$

but how to exclude 2 with fundamental math?