Answer
$$ y'=y\left( \frac{3x^2}{ x^3+1}+ \frac{4x^3}{ x^4+2}+\frac{10x^4}{x^5+3}\right).$$
Work Step by Step
Taking the $\ln $ of both sides of the equation, we get
$$\ln y= \ln((x^3+1)(x^4+2)(x^5+3)^2)$$
Then using the properties of $\ln $, we have
$$\ln y= \ln (x^3+1) +\ln (x^4+2)+2\ln(x^5+3).$$
Now taking the derivative for the above equation, we have
$$\frac{y'}{y}= \frac{3x^2}{ x^3+1}+ \frac{4x^3}{ x^4+2}+\frac{10x^4}{x^5+3},$$
Hence, $ y'$ is given by
$$ y'=y\left( \frac{3x^2}{ x^3+1}+ \frac{4x^3}{ x^4+2}+\frac{10x^4}{x^5+3}\right).$$
