If $f(xy)=f(x)f(y)$ then show that $f(x) = x^t$ for some t 
Let  $f(xy) =f(x)f(y)$ for all $x,y\geq 0$. Show that $f(x) = x^p$ for some $p$. 

I am not very experienced with proof. If we let $g(x)=\log (f(x))$ then this is the same as  $g(xy) = g(x) + g(y)$ 
I looked up the hint and it says let $g(x) = \log f(a^x) $ 
The wikipedia page for functional equations only states the form of the solutions without proof.
Attempt 
Using the hint (which was like pulling a rabbit out of the hat)
Restricting the codomain $f:(0,+\infty)\rightarrow (0,+\infty)$
so that we can define the real function $g(x) =  \log f(a^x)$ and we
have $$g(x+y) = g(x)+ g(y)$$ 
i.e $g(x) = xg(1)$ as $g(x)$ is continuous (assuming $f$ is).
Letting $\log_a f(a) = p$ we get $f(a^x) =a^p $. I do not have a rigorous argument but I think I can conclude that $f(x) = x^p$ (please fill any holes or unspecified assumptions) Different solutions are invited

Both the answers above are very good and thorough, but given an assumption that the function is differentiable, the DE approach strikes me as the easiest.
$ \frac{\partial}{\partial y} f(x y) = x f'(xy) = f(x)f'(y) $
Evaluating y at 1 gives:
$ xf'(x) = f(x)f'(1) $
The above is a separable DE:
Let $ p = f'(1) $ and $ z = f(x) $
$ x\frac{dz}{dx} = pz \implies \int \frac{dz}{z} = p\int \frac{dx}{x}$
$ \therefore \ln|z| = p\ln|x| + C $
Let $ A = e^C $. $ \implies C = \ln(A) $
$ x > 0 \implies |x| = x $
$ \therefore \ln|z| = p\ln(x) + \ln(A) = \ln(x^p) + \ln(A) = \ln(Ax^p) $
Hence $ |z| = Ax^p $; $ z = \pm Ax^p = f(x)$
Let $ B = \pm A $ and now $ f(x) = Bx^p $
Now using the initial property:
$ f(x)f(y) = Bx^p By^p = B^2 (xy)^p = f(xy) = B (xy)^p $
$B^2 = B \implies B $ is $0$ or $1$.
If B is zero, that provides the constant function $ f(x) = 0 $, otherwise the solution is $ f(x) = x^p $.
As can be seen from the other answers, this does not capture all possible solutions, but sometimes that's the price of simplicity.
