a) \(x^3+y^3=9x^3\)
\(\Leftrightarrow y^3-8x^3=0\)
\(\Leftrightarrow\left(y-2x\right)\left(y^2+2xy+4x^2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}y-2x=0\\y^2+2xy+4x^2=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}y=2x\\\left(x+y\right)^2+3x^2=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}y=2x\\\left\{{}\begin{matrix}x=0\\y=0\end{matrix}\right.\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}y=2x\\x\in N\end{matrix}\right.\)

b) We realize that (0,0,0) is a solution for the equation.
 

b) \(x^3+3y^3=9z^3\)
We realize that   is a solution for equation. We need proof the equation hasn't other solution.
Suppose if the equation have above have some other solutions then which have to exist solutions  have form (a,b,c).
By properties of natural numbers then in all such solutions, which have to solution (m,n,p) with m is the smallest natural number.
We have : \(m^3+3n^3=9p^3\)
Because  \(3n^3⋮3,9p^3⋮3\Rightarrow m^3⋮3\)​ ​\(\Leftrightarrow m=3s\)
So (1) \(\Leftrightarrow27s^3+3n^3=9p^3\)   \(\Leftrightarrow9s^3+n^3=3p^3\) (2)
The same \(n⋮3\Leftrightarrow n=3t\)
So (2) \(\Leftrightarrow9s^3+27t^3=3p^3\)\(\Leftrightarrow3s^3+9t^3=p^3\)(3)
The same \(p⋮3\Leftrightarrow p=3u\)
So (3)  \(\Leftrightarrow3s^3+9t^3=27u^3\)\(\Leftrightarrow s^3+3t^3=9u^3\)
The (4) demonstrate (s,t,u) is a solution of the equation but \(s< m\) (conflict).
So the equation have a solution is (0,0,0).