Why these are isomorphic each other for given these rings?

As Chris hinted, it is easy to lift up $\,\Bbb Z_{15} \cong \Bbb Z_3\times \Bbb Z_5\,$ by CRT. Let's examine the idea more closely.

Notice in $\,R = \Bbb Z/15\!:\ (3)+(5)=(1)\,\Rightarrow\, (3)\cap (5) = (3)(5) = (0)$ $\smash{\overset{\small\rm CRT}\Rightarrow}\, R^{\phantom{|^|}}\!\!\! \cong R/3\times R/5$

The above ideal equalities extend to $\,E = R[x]/(3x^3+5x),\,$ thus also $\smash{\overset{\small\rm CRT}\Rightarrow}\, E^{\phantom{|^|}}\!\!\! \cong E/3\times E/5$

Ring isomorphism theorems $\Rightarrow E/3 \cong \Bbb Z_3[x]/5x,\,$ $\,E/5 \cong \Bbb Z_5[x]/3x^2$

For variety here's another way: we apply CRT in $R=\Bbb Z_{\color{#c00}{15}}[x]\,$ with $\,I+J=(3,5x) + (5,3x^2) \supseteq (5,3)= (1).\,$ By ring isomorphism theorems $\,R/I = R/(3,5x)^{\phantom{|^|}}\!\!\! \cong \Bbb Z_3[x]/(5x),\,$ $\,R/J = R/(5,3x^2) \cong \Bbb Z_5[x]/(3x^2)$

$I\!+\!J=(1)\,\Rightarrow\,I\cap J = IJ = (3,5x)(5,3x^2) =(9x^2,-5x)= (3x^2,5x)\,$ by $\,2(9x^2)=3x^2$

hence $\,IJ =(3x^2,5x)=(3x^2\!+\!5x)\ $ by $\ (6,-5)(3x^2\!+\!5x)=(3x^2,5x)\,$ by $\,\color{#c00}{15=0}\,$ in $\,R$.

Conclude $\ R/(3x^2\!+\!5x) = R/(I\cap J)\overset{\rm\small CRT_{\phantom |}\!} = R/I\times R/J = \Bbb Z_3[x]/(5x)\times \Bbb Z_5[x]/(3x^2)$