Calculate the dual basis and tangent basis vectors

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A coordinate system with the coordinates s and t in $R^2$ is defined by the coordinate transformations: $s = y/y_0$ and $t=y/y_0 - tan(x/x_0)$ , where $x_0$ and $y_0$ are constants.

a) Determine the area that includes the point (x, y) = (0, 0) where the coordinate system is well defined. Express the area both in the Cartesian coordinates (x, y) and in the new coordinates (s, t).

b) Calculate the tangent basis vectors $\vec E_s$ and $\vec E_t$ and the dual basis vectors $\vec E^s$ and $\vec E^t$

c)Determine the inner products $\vec E_s\cdot\vec E^s$, $\vec E_s\cdot\vec E^t$, $\vec E_t\cdot\vec E^s$ and $\vec E_t\cdot\vec E^t$

My attempt: a) Since $tan(x/x_0)$ is not defined for $x=\pm\pi/2\cdot x_0$ I assume x must be in between those values therefore $-\pi/2\cdot x_0 < x < \pi/2\cdot x_0$ and y can be any real number. Is this the correct answer on a)?

b) I can solve x and y for s and t which gives me $y=y_0\cdot s$ and $x=x_0\cdot arctan(s-t)$. $\vec E_s = \frac {x_0} {1 + (s-t)^2}\cdot\vec e-x + y_0\cdot\vec e_y$ and $\vec E_t = - \frac { x_0} { 1 + (s-t)^2}\cdot\vec e_x$. I get the dual basis vectors from $\vec E^s = \frac {1} {y_0}\cdot\vec e_y$ and $\vec E^t = \frac {1} {y_0}\cdot\vec e_y - \frac {1} {x_0(1+(x/x_0)^2)}\cdot\vec e_x$ , is this the correct approach?

c) It was here that I really started to question if i had done correct on a and b since I get $\vec E_s\cdot \vec E^s = 1 $and$\vec E_t\cdot \vec E^s = 0$, this feels correct but then i get by just plugging in $\vec E_t\cdot \vec E^t = \frac {x_0} {(1+(s-t)^2)(1+arctan(s-t)^2)} $and $\vec E_s\cdot \vec E^t = 1-\frac {1} {(1+(s-t)^2)(1+arctan(s-t)^2)}$. Is this really correct? Because it feels like it is not correct.

Thanks in advance!