$f(1)=f(1)g(0)+f(0) > f(0) \Longrightarrow f(1)g(0) >0 \Longrightarrow f(1)\neq 0$ $f(x+1)=f(x)g(1)+f(1)=f(1)g(x)+f(x) \Longrightarrow $ $f(x)(g(1)-1)=f(1)g(x)-f(1)=f(1)(g(x)-1)$ case 1) $g(1)=1 \Longrightarrow g(x)=1\Longrightarrow f(x+y)=f(x)+f(y)$. But $f$ is increasing so $f(x)=cx $ with $c>0$ case 2) $g(1)\neq 1 \Longrightarrow f(x)=\frac {f(1)}{g(1)-1}(g(x)-1)=k(g(x)-1) \Longrightarrow $ $k(g(x+y)-1)=k(g(x)-1)g(y)+k(g(y)-1)\Longrightarrow g(x+y)=g(x)g(y)$ But $g$ is monotonic because $f$ is increasing , so $g(x)=a^x$ with $ a>0,a\neq 1$ and $f(x)=\frac c{a-1}(a^x-1)$ with $ c>0$

can someone else help us solve this exercise in an other way? please....