Given that f^prime(x)>g^prime(x) for all x ∈ R and f(0)=g(0), then

Let $h(x)=f(x)-g(x)$ for all $x \in R$. Then,

$h'(x)=f'(x)-g'(x)$ for all $x \in R$

$\Rightarrow h'(x)>0$ for all $x \in R$

$\Rightarrow h(x)$ is increasing on R.

But, $h(0)=f(0)-g(0)=0$

∴  $h(x)>h(0)$ for all $x>0$

and,

$h(x)<h(0)$ for all $x<0$

$\Rightarrow f(x)-g(x)>0$ for all $x(0, \infty)$

and,

$f(x)-g(x)<0$ for all $x \in(-\infty, 0)$

$\Rightarrow f(x)>g(x)$ for all $x \in(0, \infty)$

and,

$f(x)<g(x)$ for all $x \in(-\infty, 0)$