Problem. Find the function \(f(x)\) satisfying $$ f(x)f(y)-f(xy)-90=\frac{10(x+y)}{xy}. $$

Problem. Find the function \(f(x)\) satisfying

$$ f(x)f(y)-f(xy)-90=\frac{10(x+y)}{xy}. $$

Solution.

Rewrite the equation as $$ f(x)f(y)-f(xy)=90+\frac{10}{x}+\frac{10}{y}. $$ Assume $$ f(x)=a+\frac{b}{x}. $$ Then $$ f(y)=a+\frac{b}{y}, $$ and $$ \begin{aligned} f(x)f(y)-f(xy) &=\left(a+\frac{b}{x}\right)\left(a+\frac{b}{y}\right) -\left(a+\frac{b}{xy}\right)\\ &=a^2-a+\frac{ab}{x}+\frac{ab}{y} +\frac{b^2-b}{xy}. \end{aligned} $$ Comparing coefficients with $$ 90+\frac{10}{x}+\frac{10}{y}, $$ gives $$ a^2-a=90, $$ $$ ab=10, $$ $$ b^2-b=0. $$ From $$ b^2-b=0, $$ we obtain $$ b=0 \quad \text{or} \quad b=1. $$ Since \(ab=10\), we must have $$ b=1. $$ Then $$ a=10. $$ Therefore, $$ \boxed{f(x)=10+\frac{1}{x}.} $$ Verification: $$ \begin{aligned} f(x)f(y)-f(xy)-90 &=\left(10+\frac1x\right)\left(10+\frac1y\right) -\left(10+\frac1{xy}\right)-90\\ &=\frac{10}{x}+\frac{10}{y}\\ &=\frac{10(x+y)}{xy}, \end{aligned} $$ which satisfies the given equation.