isn't it possible to get partial derivatives for x^(1/3)y^(1/3)?

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I'm studying "vector calculus written by Jerrold E.marsden, Anthony Tromba 6th Ed"

My problem is I couldn't understand example 4 in 2.3section

They said that they can get partial derivatives of $x^{1/3}y^{1/3}$ at (0,0) like below $$f_x(x,y)=lim_{h\rightarrow0}\frac{f(h,0)-f(0,0)}{h}=lim_{h\rightarrow0}\frac{0-0}{h}=0$$

Next paragraph, they said there are no partial derivatives in $x^{1/3}y^{1/3}$ because It doesn't have tangent plane at (0,0)

I'm confusing these two contradictory explanation.

It seems like that I could get partial derivatives by definition of partial derivative.

But I know It couldn't be happen

what is the turth of this example?