I used the method of Lagrange's multpliers to find the maximum of $f(x,y,z)=\ln x+\ln y+3\ln z$ on the portion of the sphere $g(x,y,z)=x^2+y^2+z^2=5r^2 \ ; r>0$ where $x>0, y>0, z>0$ .
I found that the point $(r,r,\sqrt 3r)$ satisfies in the system $\nabla f(x,y,z)=\lambda\nabla g(x,y,z),\ g(x,y,z)=5r^2$ for some $\lambda\in\mathbb{R}$ uniquely, (Of course $\lambda=\frac{1}{2r^2}$).
Consequently, $f(r,r,\sqrt 3r)=\ln(3\sqrt3r^5)$ is only value of the solution of system.
My question is that why $\ln(3\sqrt3r^5)$ is maximum $f$ on $g=5r^2$ on the first octant.
How do I prove it algebraically?
First thing first: you can simplify that $r$ out of the problem since it is homogeneous (but I would keep the $5$ since it does simplify the problem). Second thing: $f(x) = \ln x + \ln y + 3 \ln z = \ln (xyz^3)$, and $\ln$ is monotonic, so you should discard it.
As Winther said, this is the only stationary point. Your octant is border-less and therefore non-compact: therefore, functions on it do not necessarily have a maximum, and may even be unbounded. But if you can prove that $xyz^3$ has a maximum on the "octisphere", then this maximum is a critical point, of which there is only one, $(1,1,\sqrt{3})$.
You can do this by cutting the octisphere. For fixed $z$, it is easy to see that, for $x^2 + y^2 = 5-z^2$, $xy$ is maximal for $x = y = \sqrt{(5-z^2)/2}$, the maximal value then being $xy = (5-z^2)/2$. Now it is a simple exercise to check that $z^3 (5-z^2)/2$ does have a maximum on $]0,1]$, and even to find it.