I have taken the exp on both sides to give;
$$e^{\ln(1+y)}-e^{\ln y}=e^{\ln(x^3+1)}-e^{2\ln(x-1)}$$ $$1+y-y=x^3+1-(x-1)^2$$
Then simplified the LHS and factored the RHS to give;
$$y=(x-1)(x^2-2)$$
Is this correct, or have i gone too far down the rabbit hole?
Taking the exponent is not linear: $$a+b=c+d$$ Means: $$\exp(a+b)=\exp(c+d) $$ $$\exp(a) \exp(b) =\exp(c) \exp(d) $$
My hint to you is to first combine the logarithms using the identities: $$ \log(a)+\log(b) = \log(ab)$$ $$ \log(a)-\log(b) = \log(\frac{a}{b})$$ $$c \log(a) = \log(a^c) $$
If you use these, you can simplify the terms to two logs on each side and the apply the exponentiation. You will get something like: $$\ln\left(\frac{1+y}{y}\right)= \ln\left(\frac{x^3+1}{(x-1)^2}\right)$$ $$\ln\left(1+\frac{1}{y}\right)= \ln\left(\frac{x^3+1}{(x-1)^2}\right)$$ So: $$1+\frac{1}{y} =\frac{x^3+1}{(x-1)^2}$$