The question simply asks to find the solution of differential equations $\frac{dy}{dx} = 2y+z$ and $\frac{dz}{dx}=3y$ that satisfy the relation:
a) $(y + z)e^{3x} = k$
b) $(3y + z)e^{-3x} = k$
c) $(3y + z)e^{3x} = k$
d) $(y - z)e^{3x} = k$
I began by integrating the second differential equation $w.r.t.x$ and got
$$ z = 3xy$$
and substituted this into the first equation and integrated again $\text{w.r.t. } x$ $$y = 2yx + \frac{3yx^{2}}{z}$$
I do not know how to proceed from here.
NB: This isn't a homework question. Practicing for an examination
We have
$$y''-2y'-3y=0\implies y=Ae^{-x}+Be^{3x}$$
$$\frac{dz}{dx}=3y=3Ae^{-x}+3Be^{3x}\implies z=-3Ae^{-x}+Be^{3x}+C$$
Then
$$3y+z=4Be^{3x}+C\implies(3y+z)e^{-3x}=k$$
where $k=4B+Ce^{-3x}$