Personally, I would not have solved for y, taking logarithms to begin with. (Perhaps I just don't like logarithms!)

I would have first determined that, when x= 1, we have \(\displaystyle e^{y}= \frac{1(1+ 1)^3}{1^2+ 1}= \frac{8}{2}= 4\) so that y= ln(4).

Then, using "implicit differentiation" \(\displaystyle e^{xy}+ e^{xy}y'= \frac{(x+1)^3}{x^2+ 1}+ \frac{3x(x+1)^2}{x^2+ 1}- \frac{2x^2(x+ 1)^3}{(x^2+ 1)^2}\) so that, with x= 1, y= 4,

\(\displaystyle e^4+ e^4y'= \frac{(1+ 1)^3}{1+ 1}+ \frac{3(1)(1+1)^2}{1+ 1}- \frac{2(1)(1+1)^3}{(1+ 1)^2}\)

\(\displaystyle e^4(1+ y')= \frac{8}{2}+ \frac{3(4)}{2}- \frac{2(8)}{4}= 4+ 6- 4= 6\)

so that \(\displaystyle 1+ y'= 6e^{-4}\) and \(\displaystyle y'= 6e^{-4}- 1\).

The equation of the tangent line is \(\displaystyle y= (6e^{-4}-1)(x- 1)+ ln(4)\).