Rewrite the function of the graph to vertex form, we have
$\begin{array}{rcl}
f(x) & = & 3x^2+18mx+22m^2 \\
f(x) & = & 3(x^2+6mx)+22m^2 \\
f(x) & = & 3\left[x^2+6mx +\left(\dfrac{6m}{2}
\right)^2-\left(\dfrac{6m}{2}\right)^2\right]+22m^2 \\
f(x) & = & 3[(x+3m)^2-9m^2]+22m^2\\
f(x) & = & 3(x+3m)^2-5m^2
\end{array}$
Hence, we have
$\begin{array}{rcl}
y & = & -f(3x) \\
y & = & -[3(3x+3m)^2-5m^2] \\
y & = & -27(x+m)^2+5m^2 \ \ldots\unicode{x2460}
\end{array}$
I may not be true. From $\unicode{x2460}$, the $x$-coordinate of the vertex of the graph is $-m$. The statement is true only when $m=0$.
II must be true. From $\unicode{x2460}$, the $y$-coordinate of the vertex of the graph is $5m^2$.
III must be true. From $\unicode{x2460}$, the equation of the axis of symmetry of the graph is $x=-m$, i.e. $x+m=0$.
