Consider two bases $e_i$ and $f_i$ of $\mathbb{R}^2$ defined by: $$\begin{aligned} (f_1,f_2) &= (e_1,e_2)\cdot F\\ \begin{pmatrix}1&-1\\1&1\end{pmatrix} &= \begin{pmatrix}1&0\\0&1\end{pmatrix} \cdot \begin{pmatrix}1&-1\\1&1\end{pmatrix} \end{aligned}$$
and the cooresponding change of coordinates ($B\equiv F^{-1}$): $$\begin{aligned} B\cdot \begin{pmatrix}x\\y\end{pmatrix} &= \begin{pmatrix}\tilde x\\ \tilde y \end{pmatrix}_f = \begin{pmatrix}1/2 x + 1/2 y \\ -1/2 x+1/2y\end{pmatrix}_f \\ F\cdot \begin{pmatrix}\tilde x\\ \tilde y \end{pmatrix}_f &= \begin{pmatrix}x\\y\end{pmatrix} = \begin{pmatrix}\tilde x-\tilde y\\ \tilde x+\tilde y\end{pmatrix} \end{aligned}$$
I would like to find the gradient of a function in both coordinate systems. Here, tilde symbols are associated to the $f$-basis. Consider $$\begin{aligned} V(x,y) &= x^2 +y \\ \tilde V(\tilde x,\tilde y) &= (\tilde x-\tilde y)^2 + (\tilde x + \tilde y) \end{aligned}$$
The gradients are defined using the metric-tensor scalings, as $$\begin{aligned} \nabla &= \partial_x e_1 + \partial_y e_2\\ \tilde \nabla &= \frac{1}{\sqrt{f_1\cdot f_1}} \partial_{\tilde x} f_1 +\frac{1}{\sqrt{f_2\cdot f_2}} \partial_{\tilde y} f_2 = \frac{1}{\sqrt{2}} \partial_{\tilde x} f_1 +\frac{1}{\sqrt{2}} \partial_{\tilde y} f_2 \end{aligned}$$
Lets check the gradients at the point $(2,1)\equiv (1.5, -0.5)_f$ where the second coordinates are with respect to the $f$-basis:
$$\begin{aligned} \nabla V \vert_{(2,1)} &= \begin{pmatrix}2x \\1 \end{pmatrix}= \begin{pmatrix}4\\ 1\end{pmatrix} \\ \tilde \nabla\tilde V \vert_{(1.5,-0.5)}&= \frac{1}{\sqrt{2}} [2(\tilde x-\tilde y)+1] f_1 + \frac{1}{\sqrt{2}} [-2(\tilde x-\tilde y)+1] f_2 \\ &= \frac{1}{\sqrt{2}} 5 \begin{pmatrix}1\\1 \end{pmatrix} + \frac{1}{\sqrt{2}} (-3) \begin{pmatrix}-1\\1 \end{pmatrix} \\ &= \frac{1}{\sqrt{2}} \begin{pmatrix}8\\2 \end{pmatrix} \end{aligned}$$
Why do i get a different gradient vector as a result?