[tex]\frac{dy}{dy}=1[/tex], so I assume you mean "find [tex]\frac{dy}{dx}[/tex]".
We can rewrite this as an implicit equation to avoid using too much of the chain rule, namely
[tex]y = \sqrt[3]{\dfrac{e^x (x+1)}{x^2+1}} \implies (x^2+1) y^3 = e^x (x+1)[/tex]
Differentiate both sides with respect to [tex]x[/tex] using the product and chain rules.
[tex]2x y^3 + 3(x^2+1) y^2 \dfrac{dy}{dx} = e^x(x+1) + e^x[/tex]
[tex]\implies 3(x^2+1) y^2 \dfrac{dy}{dx} = e^x (x+2) - 2x y^3[/tex]
[tex]\implies \dfrac{dy}{dx} = \dfrac{e^x (x+2) - 2x y^3}{3(x^2+1) y^2}[/tex]
Now substitute the original expression for [tex]y[/tex].
[tex]\dfrac{dy}{dx} = \dfrac{e^x (x+2) - 2x \left(\sqrt[3]{\frac{e^x(x+1)}{x^2+1}}\right)^3}{3(x^2+1) \left(\sqrt[3]{\frac{e^x(x+1)}{x^2+1}}\right)^2}[/tex]
[tex]\implies \dfrac{dy}{dx} = \dfrac{e^x (x+2) - \frac{2e^x(x^2+x)}{x^2+1}}{3(x^2+1) \left(\frac{e^x(x+1)}{x^2+1}\right)^{2/3}}[/tex]
[tex]\implies \dfrac{dy}{dx} = e^x \dfrac{x^3-x+2}{3(x^2+1)^2 \frac{e^{2x/3}(x+1)^{2/3}}{(x^2+1)^{2/3}}}}[/tex]
[tex]\implies \dfrac{dy}{dx} = e^{x/3} \dfrac{x^3-x+2}{3(x^2+1)^{4/3} (x+1)^{2/3}}[/tex]
Now, since
[tex]y = \sqrt[3]{\dfrac{e^x (x+1)}{x^2+1}} = \dfrac{e^{x/3} (x+1)^{1/3}}{(x^2+1)^{1/3}}[/tex]
we can write
[tex]\dfrac{dy}{dx} = e^{x/3} \dfrac{x^3-x+2}{3(x^2+1)^{4/3} (x+1)^{2/3}} = \dfrac{e^{x/3} (x+1)^{1/3}}{(x^2+1)^{1/3}} \times \dfrac{x^3-x+2}{3(x^2+1)^{3/3} (x+1)^{3/3}}[/tex]
[tex]\implies \dfrac{dy}{dx} = y \dfrac{x^3-x+2}{3(x^2+1)(x+1)}[/tex]
Focusing on the rational expression in [tex]x[/tex], we have the partial fraction expansion
[tex]\dfrac{x^3 - x + 2}{(x^2 + 1) (x+1)} = a + \dfrac{bx+c}{x^2+1} + \dfrac d{x+1}[/tex]
where we have the constant term on the right side because both the numerator and denominator have degree 3.
Writing everything with a common denominator and equating the numerators leads to
[tex]x^3 - x + 2 = a (x^2+1) (x+1) + (bx+c)(x+1) + d(x^2+1) \\\\ = ax^3 + (a+b+d)x^2 + (a+b+c)x + a+c+d[/tex]
[tex]\implies \begin{cases} a = 1 \\ a+b+d=0 \\ a+b+c = -1 \\ a+c+d=2 \end{cases}[/tex]
[tex]\implies a=1, b=-2, c=0, d=1[/tex]
[tex]\implies \dfrac{x^3 - x + 2}{(x^2 + 1) (x+1)} = 1 - \dfrac{2x}{x^2+1} + \dfrac 1{x+1}[/tex]
and it follows that
[tex]\boxed{\dfrac{dy}{dx} = \dfrac y3 \left(1 - \dfrac{2x}{x^2+1} + \dfrac1{x+1}\right)}[/tex]