f(x)f(y) − f(xy) = x + y

Step 1. Put $y=0$: $c\,f(x)-c=x$, i.e. $c(f(x)-1)=x$ for all $x$. If $c=0$, this gives $x=0$ for all $x$, absurd. So $c\ne0$ and
$$f(x)=1+\frac xc.$$

Step 2. Substitute into the equation:
$$\left(1+\frac xc\right)\left(1+\frac yc\right)-1-\frac{xy}c=x+y,$$
which simplifies to
$$(x+y)\left(\frac1c-1\right)+xy\left(\frac1{c^2}-\frac1c\right)=0\qquad\text{for all }x,y.$$
Taking $y=0$, $x=1$: $\frac1c-1=0$, so $c=1$. (Then the second coefficient is $1-1=0$ as well.)

Answer. $f(x)=x+1$. Check: $(x+1)(y+1)-(xy+1)=x+y$ ✓.

Lesson. Substituting $0$ for a variable often makes the equation linear in $f(x)$: always try it first, and remember to exclude the degenerate value ($c=0$) before dividing.