Answer
$ a.\quad +\infty$
$ b.\quad +\infty$
$ c.\quad +\infty$
$ d.\quad +\infty$
Work Step by Step
$ a.\qquad x\rightarrow 0^{+}\Rightarrow$
$\left[\begin{array}{ll}
& x^{2/3}=(x^{1/3})^{2}\rightarrow 0^{+}\\
& \\
& \frac{1}{x^{2/3}}\rightarrow+\infty
\end{array}\right],\ \ \left[\begin{array}{l}
(x-1)^{2/3}=[(x-1)^{1/3}]^{2}\rightarrow+1\\
\\
\frac{2}{(x-1)^{2/3}}\rightarrow+2
\end{array}\right], $
$\displaystyle \Rightarrow(\frac{1}{x^{2/3}}+\frac{2}{(x-1)^{2/3}})\rightarrow+\infty$
$ b.\qquad x\rightarrow 0^{-}\Rightarrow$
$\left[\begin{array}{ll}
& x^{2/3}=(x^{1/3})^{2}\rightarrow 0^{+}\\
& \\
& \frac{1}{x^{2/3}}\rightarrow+\infty
\end{array}\right],\ \ \left[\begin{array}{l}
(x-1)^{2/3}=[(x-1)^{1/3}]^{2}\rightarrow+1\\
\\
\frac{2}{(x-1)^{2/3}}\rightarrow+2
\end{array}\right], $
$\displaystyle \Rightarrow(\frac{1}{x^{2/3}}+\frac{2}{(x-1)^{2/3}})\rightarrow+\infty$
$ c.\qquad x\rightarrow 1^{+}\Rightarrow$
$\left[\begin{array}{ll}
& x^{2/3}=(x^{1/3})^{2}\rightarrow+1\\
& \\
& \frac{1}{x^{2/3}}\rightarrow 1
\end{array}\right],\ \ \left[\begin{array}{l}
(x-1)\rightarrow 0^{+}\\
\\
(x-1)^{2/3}=[(x-1)^{1/3}]^{2}\rightarrow+0\\
\\
\frac{2}{(x-1)^{2/3}}\rightarrow+\infty
\end{array}\right], $
$\displaystyle \Rightarrow(\frac{1}{x^{2/3}}+\frac{2}{(x-1)^{2/3}})\rightarrow+\infty$
$ d.\qquad x\rightarrow 1^{-}\Rightarrow$
$\left[\begin{array}{ll}
& x^{2/3}=(x^{1/3})^{2}\rightarrow+1\\
& \\
& \frac{1}{x^{2/3}}\rightarrow 1
\end{array}\right],\ \ \left[\begin{array}{l}
(x-1)\rightarrow 0^{-}\\
\\
(x-1)^{2/3}=[(x-1)^{1/3}]^{2}\rightarrow 0^{+}\\
\\
\frac{2}{(x-1)^{2/3}}\rightarrow+\infty
\end{array}\right], $
$\displaystyle \Rightarrow(\frac{1}{x^{2/3}}+\frac{2}{(x-1)^{2/3}})\rightarrow+\infty$
