$y^{y\sqrt[3]y} = (y\sqrt[3]y)^y$

$\begin{aligned} y^{y\sqrt[3]y} & = (y\sqrt[3]y)^y \\ y^{y^\frac 43} & = y^{\frac 43y} & \small \color{#3D99F6} \text{Taking log on both sides} \\ y^\frac 43 \ln y & = \frac 43 y \ln y \\ y^\frac 43 \ln y - \frac 43 y \ln y & = 0 \\ y \left(y^\frac 13 - \frac 43\right) \ln y & = 0 \end{aligned}$

$\implies y = \begin{cases} y = 0 & \small \color{#D61F06} \text{Rejected as }0^0 \text{ is not defined.} \\ y^\frac 13 - \dfrac 43 = 0 & \implies y = \dfrac {64}{27} \\ \ln y = 0 & \implies y = 1 \end{cases}$

Therefore, the sum of all possible values of $y$ is $\dfrac {64}{27}+1 = \dfrac {91}{27}$ and $a+b = 91 + 27 = \boxed{118}$ .

$\large \begin{aligned} y^{y\sqrt[3]y} & = (y\sqrt[3]y)^y \\ \Rightarrow (y^y)^{\sqrt[3]y} & = (y \cdot y^{\frac 13})^y \\ \Rightarrow (y^y)^{\sqrt[3]y} & = (y^{\frac 43})^y \\ \Rightarrow (y^y)^{\sqrt[3]y} & = (y^y)^{\frac 43} \\ \Rightarrow \sqrt[3]y & = \dfrac 43 \\ \implies y & = \dfrac{64}{27} \\ \end{aligned}$

Since $1^n = 1$ for any number $n,$ the other solution must be $y = 1$ . So $\dfrac ab = \dfrac{64}{27} +1 = \dfrac{91}{27}$ . Hence $a + b = 91 + 27 = \boxed{118}$