Using the power series method
$$ y = \sum_{n=0}^\infty c_nx^n, \quad xy' = \sum_{n=0}^\infty nc_n x^n $$
$$ y'' = \sum_{n=2} n(n-1)c_nx^n = \sum_{n=0}^\infty (n+2)(n+1)c_{n+2}x^n $$
Plugging these into the equation, we find a recurrence relation
$$ (n+2)(n+1)c_{n+2} + 3(n-2)c_n = 0 $$
Clearly, the odd and even coefficients are separated from each other. The even series is finite, with
$$ c_2 = 3c_0, c_{n>2} = 0 $$
Therefore, we can write
$$ y(x) = c_0 (1+3x^2) + c_1 y_2(x) $$
where $y_2(x)$ is the remaining odd power series with $y_2(0) = 0$ and ${y_2}'(0)=1$
We can derive a closed form expression for the coefficients, but a better method would be to use reduction of order. Substitute $y_2(x) = (1+3x^2)u(x)$ to get a new equation
$$ u'' + \left(3x + \frac{12x}{1+3x^2} \right)u' = 0 $$
where
$$ u(0) = 0, \quad u'(0) = 1 $$
Integrating and applying the above conditions, we find
$$ u'(x) = \frac{e^{-3x^2/2}}{(1+3x^2)^2}, \quad u(x) = \int_0^x \frac{e^{-3t^2/2}}{(1+3t^2)^2} dt $$
The last step is to determine the free constants. We apply the given B.C.s
\begin{align} y(0) &= c_0 = 1 \\ y'(1) &= 6 + c_1 {y_2}'(1) = 0.1 \end{align}
where numerical integration must be performed to find
$$ {y_2}'(1) = 4u'(1) + 6u(1) = \frac{e^{-3/2}}{4} + 6\int_0^1\frac{e^{-3t^2/2}}{(1+3t^2)^2} dt \approx 2.213185 $$
$$ \implies c_1 = -\frac{5.9}{{y_2}'(1)} \approx -2.665481 $$