Answer
$1-\dfrac{1}{2 \cdot 2!}x^{4/3}+\dfrac{1}{2^2 \cdot 4!}x^{8/3}-\dfrac{1}{2^3 \cdot 6!}x^{4}+..$
Work Step by Step
Consider the Taylor's Series for $\cos x$ as follows:
$ \cos x=1-\dfrac{x^2}{2!}+\dfrac{x^4}{4!}-\dfrac{x^6}{6!}+...$
Plug $\dfrac{x^{2/3}}{\sqrt 2}$ instead of $x$ in the above series as follows:
Now, $ \cos(\dfrac{x^{2/3}}{\sqrt 2})=1-\dfrac{(\dfrac{x^{2/3}}{\sqrt 2})^2}{2!}+\dfrac{(\dfrac{x^{2/3}}{\sqrt 2})^4}{4!}-\dfrac{(\dfrac{x^{2/3}}{\sqrt 2})^6}{6!}+...$
$\implies 1-\dfrac{1}{2 \cdot 2!}x^{4/3}+\dfrac{1}{2^2 \cdot 4!}x^{8/3}-\dfrac{1}{2^3 \cdot 6!}x^{4}+..$
