Answer
$f(g(x))=x^2-3$
$g(f(x))=x^2-6x+9$
Work Step by Step
f of g(x) means $f(g(x))$
$g(x)$ is given as $x^2$
Therefore $f(g(x))=f(x^2)=x^2-3$
g of f(x) means $g(f(x))$
$f(x)$ is given as $x-3$
So $g(f(x))=g(x-3)=(x-3)^2$
By expanding $(x-3)^2$, $g(f(x))=x^2-6x+9$