Answer
The graph of $y=4x^{2/5}-2x$ does not have any vertical tangents anywhere.
Work Step by Step
$$y=f(x)=4x^{2/5}-2x$$
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:
$$f′(0)=\lim_{h\to0}\frac{f(h)−f(0)}{h}=\lim_{h\to0}\frac{(4h^{2/5}-2h)-(4\times0^{2/5}-2\times0)}{h}$$
$$f'(0)=\lim_{h\to0}\frac{4h^{2/5}-2h}{h}=\lim_{h\to0}\Big(\frac{4}{h^{3/5}}-2\Big)$$
As $h\to0^+$, $h^{3/5}\to0^+$, so $\frac{4}{h^{3/5}}\to\infty$ and $\lim_{h\to0^+}(\frac{4}{h^{3/5}}-2)=\infty$
As $h\to0^-$, $h^{3/5}\to0^-$, so $\frac{4}{h^{3/5}}\to-\infty$ and $\lim_{h\to0^-}(\frac{4}{h^{3/5}}-2)=-\infty$
This means $\lim_{h\to0}(\frac{4}{h^{3/5}}-2)$ does not exist. Therefore, as predicted, $f$ does not have a vertical tangent at the origin.