For [tex]0\le t\le1[/tex], we have
[tex]y'+6y=1\implies e^{6t}y'+6e^{6t}=(e^{6t}y)'=e^{6t}\implies y=\dfrac16+Ce^{-6t}[/tex]
Given that [tex]y(0)=0[/tex], we have
[tex]0=\dfrac16+C\implies C=-\dfrac16[/tex]
so that
[tex]y=\dfrac16(1-e^{-6t})[/tex]
For [tex]t>1[/tex] (I think you mistakenly wrote [tex]t>0[/tex], which overlaps with the first definition of [tex]g(t)[/tex]), we have
[tex]y'+6y=0\implies e^{6t}y'+6e^{6t}y=(e^{6t}y)'=0\implies y=Ke^{-6t}[/tex]
We want this to be a continuation of the previously found solution [tex]y[/tex] at [tex]t=1[/tex], which means we need to pick [tex]K[/tex] such that
[tex]\dfrac16(1-e^{-6})=Ke^{-6}\implies K=\dfrac16(e^6-1)[/tex]
Then the solution to the IVP is
[tex]y(t)=\begin{cases}\frac16(1-e^{-6t})&\text{for }0\le t\le1\\\frac{e^6-1}6e^{-6t}&\text{for }t>1\end{cases}[/tex]
Alternatively, we can get around treating [tex]g(t)[/tex] piecemeal and resorting to the Laplace transform. Write
[tex]g(t)=\begin{cases}1&\text{for }0\le t\le1\\0&\text{for }t>1\end{cases}=u(t)-u(t-1)[/tex]
where
[tex]u(t-c)=\begin{cases}0&\text{for }t<c\\1&\text{for }t\ge c\end{cases}[/tex]
is the unit step function.
Take the Laplace transform of both sides of the ODE:
[tex]y'+6y=g(t)\overset{\text{L.T.}}{\implies}(sY-y(0))+6Y=\mathcal L_s\{g(t)\}[/tex]
where [tex]Y=Y(s)[/tex] is the Laplace transform of [tex]y(t)[/tex].
We have
[tex]\mathcal L_s\{g(t)\}=\displaystyle\int_0^\infty g(t)e^{-st}\,\mathrm dt=\int_0^1e^{-st}\,\mathrm dt=\dfrac{1-e^{-s}}s[/tex]
so that
[tex](s+6)Y=\frac{1-e^{-s}}s\implies Y=\dfrac{1-e^{-s}}{s(s+6)}=\dfrac{1-e^{-s}}6\left(\dfrac1s-\dfrac1{s+6}\right)[/tex]
Taking the inverse transform yields
[tex]y=\dfrac{1-u(t-1)}6-\dfrac{e^{-6t}}6(e^tu(t-1)-1)[/tex]
[tex]y=\dfrac{1-e^{-6t}}6+\dfrac{e^{6-6t}-1}6u(t-1)[/tex]
which is equivalent to the same solution found earlier; for [tex]0\le t\le1[/tex], [tex]u(t-1)=0[/tex], so that [tex]y=\frac{1-e^{-6t}}6[/tex]; for [tex]t>1[/tex], [tex]u(t-1)=1[/tex], and [tex]y=\frac{1-e^{-6t}}6+\frac{e^{6-6t}-1}6=\frac{(e^6-1)e^{-6t}}6[/tex].
