Answer
The graph of $y=x^{2/5}$ does not have any vertical tangents anywhere.
Work Step by Step
$$y=f(x)=x^{2/5}$$
a) The graph is enclosed below. It does not seem to have any vertical tangents, even at the origin, because as we see from the introduction that the graph of $y=x^{2/3}$ does not have a vertical tangent at the origin.
b) The graph of $f(x)$ has a vertical tangent at $x$ when
1) $f$ is continuous at $x$
2) $$f'(x)=\lim_{h\to0}\frac{f(x+h)-f(x)}{h}=\pm\infty$$
Here, we can see that: $$f'(0)=\lim_{h\to0}\frac{f(h)-f(0)}{h}=\lim_{h\to0}\frac{h^{2/5}-0^{2/5}}{h}=\lim_{h\to0}\frac{h^{2/5}}{h}$$
$$f'(0)=\lim_{h\to0}h^{-3/5}=\lim_{h\to0}\frac{1}{h^{3/5}}$$
As $h\to0^+$, $h^{3/5}\to0^+$, so $\lim_{h\to0^+}(1/h^{3/5})=\infty$
As $h\to0^-$, $h^{3/5}\to0^-$, so $\lim_{h\to0^-}(1/h^{3/5})=-\infty$
This means $\lim_{h\to0}(1/h^{3/5})$ does not exist. Therefore, as predicted, $f$ does not have a vertical tangent at the origin.