Functions, anyone?

Let $a=x-y$ and $b=x+y$ .

Thus $x=\frac{b+a}{2},y=\frac{b-a}{2}$ $af(b)-bf(a)=4(\frac{b+a}{2})(\frac{b-a}{2})ab$ $af(b)-bf(a)=(b+a)(b-a)ab$

Divide by both sides by $ab$ : $\frac{f(b)}{b}-\frac{f(a)}{a}=b^2-a^2$

Let $g(x)=\frac{f(x)}{x}$ : $g(b)-g(a)=b^2-a^2$ $g(b)-b^2=g(a)-a^2$

Using $g(1)=\frac{f(1)}{1}=2$ , let $a=1$ and $b=x$ : $g(x)-x^2=2-1^2$ $g(x)=x^2+1$ $f(x)=x*g(x)=x^3+x$

Hence $f(x)=\boxed{32800}$

Substitute $x=16.5, y=15.5$ in the given functional relation to obtain $f(32)=32 f(1) + 4\times 16.5 \times 15.5 \times (16.5^2-15.5^2)$ which gives $f(32)=32800$ .

Input $x=y+1$

$\begin{aligned}f(2y+1)-2(2y+1)&=4y(y+1)(2y+1)\\f(2y+1)&=(2y+1)(4{y}^2+4y+2)\\&=(2y+1)\left({(2y+1)}^2+1\right)\end{aligned}$

Input $2y+1=x$

$\begin{aligned} f(x)&=x\left({x}^2+1\right)\\f(32)&=32(1024+1)\\&\huge\color{#3D99F6}{=\boxed{32800}}\end{aligned}$