1. The only factor of $|d| = 1$ is $1$ and the factors of $|a| = 3$ are $1$ and $3$.
Taking all possible ratios $\pm\dfrac{\text{Factor of }|d|}{\text{Factor of }|a|}$ gives
$x=\pm\dfrac{\text{Factor of }1}{\text{Factor of }3} = \pm\dfrac{1}{1}$ or $\pm\dfrac{1}{3}$
$x = 1:$ \t \gap[10] $3(1)^3+5(1)^2-5(1)+1 = 4 \neq 0 $ ✘
\\ $x = -1:$ \t \gap[10] $3(-1)^3+5(-1)^2-5(-1)+1 = 8 \neq 0 $ ✘
\\ $x = \frac{1}{3}:$ \t \gap[10] $3\left(\frac{1}{3}\right)^3+5\left(\frac{1}{3}\right)^2-5\left(\frac{1}{3}\right)+1 = 0 $ ✔
Thus, $x = \frac{1}{3}$ is one solution.
2. $\ \left(x-\frac{1}{3}\right)$ is therefore a factor of the left-hand side $3x^3+5x^2-5x+1.$
To get other solutions (if any), we first factor the left-hand side using the formula above:
$ax^3+bx^2+cx+d$ \t ${}= (x-r)(ax^2 \ + \ [ar+b]x \ + \ [ar^2+br+c])$
\\ $3x^3+5x^2-5x+1$ \t ${}= \left(x-\frac{1}{3}\right)\left(3x^2 \ + \ \left\[3\left(\frac{1}{3}\right)+5\right\]x \ + \ \left\[3\left(\frac{1}{3}\right)^2+5\left(\frac{1}{3}\right)-5\right\]\right)$
\\ \t ${}= \left(x-\frac{1}{3}\right)\left(3x^2 \ + \ [1+5]x \ + \ \left\[\frac{1}{3}+\frac{5}{3}-5\right\]\right)$
\\ \t ${}= \left(x-\frac{1}{3}\right)(3x^2 \ + \ 6x \ - \ 3)$
3. The second factor is
$3x^2 + 6x - 3 = 3(x^2+2x-1).$
Although $x^2+2x-1$ does not factor over the integers, it does have two real solutions given by the quadratic formula:
$x = \dfrac{-b\pm\sqrt{b^2-4ac}}{2a} = \dfrac{-2\pm\sqrt{8}}{2} = -1 \pm \sqrt{2}.$