Draw the graph of `x^(2//3)y^(2//3)=1`

Explanation

We have `x^(2//3)+ y^(2/3)=1`
or `y^(2//3)=1-x^(2//3)`
or `y=f(x)=(1-x^(2//3))_(3//2)`
Clearly, the above function is defined if `1-x^(2//3) ge0`
or `x^(2/3) le1` or `x^(2) le1` or`-1 lexle1`
`f(0)=1`
`f(x)=0 therefore x=+-1`
`f^(')(x) =3/2(1-x^(2/3))(-2/3x^(-1//3))`
`=-(1-x^(2//3))^(1//2)/(x^(1//3))`
`f^(')(x) lt 0` for `x gt0` and `f^(')(x) lt 0` for `x lt 0`
`f^(')(0)` does not exist, sothe function is not differentiable at `x=0`.
`f^('')(x) = -((-2/3x^(-1//3))/(2(1-x^(2//3))^(1//2).x^(1//3)-1/3x^(-2//3). (1-x^(2//3))^(1//2)))/x^(2//3)`
`=1/3. (1+x^(-2//3).(1-x^(2//3)))/((1-x^(2//3))^(1//2).x^(2/3) ) lgt 0, AAx in(-1,1)`
Hence the graph is concave upward.
From the above information the graph of the function `y=f(x)` can be drawn as shown in the following figure.

For `y lt 0`, we can reflect the above graph in the x-axis, so the graph of the given relation can be drawn as shown in the following figure.