We claim that the only solution is $f(x) = 1 + \frac{1}{x}$. It's easy to see that this works. Note that $f(x) - f(xf(y)) = \frac{1}{xyf(y)}$. For any $y,z$ we have $f(x) - f(xf(y)f(z)) = f(x) - f(xf(y)) + f(xf(y)) - f(xf(y)f(z)) = \frac{1}{xyf(y)} + \frac{1}{xf(y)zf(z)}$ however the LHS is symmetric in $y,z$. Swap $y,z$ and comparing gives $(1-f(y))y = (1-f(z))z$, so $f(x) = 1-\frac{c}{x}$. Letting $x=y=1$ in the original equation gives $c = -1$, as desired.

The same as Czech 2011 P6. https://artofproblemsolving.com/community/c6h423928p2398336