We are given the component of the metric $g_{\alpha\beta}$ in a particular set of coordinates: $\{x^\gamma\} = \{y, \theta, \phi_1, \phi_2\}$. For a different set of coordinates, $\{x^{\bar{\sigma}}\}=\{R, \psi, \phi_1, \phi_2\}$, the metric has components $g_{\bar{\mu}\bar{\nu}}$ given by the transformation law between components of $0 \choose 2$ tensors.

$$g_{\bar{\mu}\bar{\nu}} = \frac{\partial x^\alpha}{\partial x^{\bar{\mu}}}\frac{\partial x^\beta}{\partial x^{\bar{\nu}}}g_{\alpha\beta}$$

You can get these partial derivatives by taking derivatives of the "substitution" in the OP. This isn't a particularly clever way of getting the new metric components, but it's the most algorithmic.

For example, differentiating both sides of the first coordinate transformation with respect to $y$

$$2\cosh{R}\sinh{R}\frac{\partial R}{\partial y} = \frac{2yh^2}{b^2} \;\;\;\;\Rightarrow \;\;\;\;\frac{\partial R}{\partial y} = \frac{2yh^2}{b^2}\frac{1}{\sinh{2R}}$$

Similarly we may obtain $\partial R/\partial \theta$, $\partial \psi/\partial y$, and $\partial \psi/\partial \theta$ and construct a matrix with elements $\partial x^{\bar{\mu}}/\partial x^\alpha$. Recall

$$\frac{\partial x^\alpha}{\partial x^{\bar{\mu}}}\frac{\partial x^{\bar{\mu}}}{\partial x^\beta} = \delta^\alpha_\beta$$

So the partial derivatives of $\{R,\psi\}$ with respect to $\{y,\theta\}$ may be obtained by inverting this matrix. Careful here we're mixing two different coordinate systems when doing these calculations as a shorthand, but eventually we should convert everything to the new coordinate system.

We are given the components of the metric $g_{\alpha\beta}$ in a particular set of coordinates: $\{x^\gamma\} = \{y, \theta, \phi_1, \phi_2\}$. For a different set of coordinates, $\{x^{\bar{\sigma}}\}=\{R, \psi, \phi_1, \phi_2\}$, the metric has components $g_{\bar{\mu}\bar{\nu}}$ given by the transformation law between components of $0 \choose 2$ tensors.

$$g_{\bar{\mu}\bar{\nu}} = \frac{\partial x^\alpha}{\partial x^{\bar{\mu}}}\frac{\partial x^\beta}{\partial x^{\bar{\nu}}}g_{\alpha\beta}$$

You can get these partial derivatives by taking derivatives of the "substitution" in the OP. This isn't a particularly clever way of getting the new metric components, but it's the most algorithmic.

For example, differentiating both sides of the first coordinate transformation with respect to $y$

$$2\cosh{R}\sinh{R}\frac{\partial R}{\partial y} = \frac{2yh^2}{b^2} \;\;\;\;\Rightarrow \;\;\;\;\frac{\partial R}{\partial y} = \frac{2yh^2}{b^2}\frac{1}{\sinh{2R}}$$

Similarly we may obtain $\partial R/\partial \theta$, $\partial \psi/\partial y$, and $\partial \psi/\partial \theta$ and construct a matrix with elements $\partial x^{\bar{\mu}}/\partial x^\alpha$. Recall

$$\frac{\partial x^\alpha}{\partial x^{\bar{\mu}}}\frac{\partial x^{\bar{\mu}}}{\partial x^\beta} = \delta^\alpha_\beta$$

So the partial derivatives of $\{y,\theta\}$ with respect to $\{R,\psi\}$ may be obtained by inverting this matrix. Careful here we're mixing two different coordinate systems when doing these calculations as a shorthand, but eventually we should convert everything to the new coordinate system.

We are given the component of the metric $g_{\alpha\beta}$ in a particular set of coordinates: $\{x^\gamma\} = \{y, \theta, \phi_1, \phi_2\}$. For a different set of coordinates, $\{x^{\bar{\sigma}}\}=\{R, \psi, \phi_1, \phi_2\}$, the metric has components $g_{\bar{\mu}\bar{\nu}}$ given by the transformation law between components of $0 \choose 2$ tensors.

$$g_{\bar{\mu}\bar{\nu}} = \frac{\partial x^\alpha}{\partial x^{\bar{\mu}}}\frac{\partial x^\beta}{\partial x^{\bar{\nu}}}g_{\alpha\beta}$$

You can get these partial derivatives by taking derivatives of the "substitution" in the OP. This isn't a particularly clever way of getting the new metric components, but it's the most algorithmic.

We are given the component of the metric $g_{\alpha\beta}$ in a particular set of coordinates: $\{x^\gamma\} = \{y, \theta, \phi_1, \phi_2\}$. For a different set of coordinates, $\{x^{\bar{\sigma}}\}=\{R, \psi, \phi_1, \phi_2\}$, the metric has components $g_{\bar{\mu}\bar{\nu}}$ given by the transformation law between components of $0 \choose 2$ tensors.

$$g_{\bar{\mu}\bar{\nu}} = \frac{\partial x^\alpha}{\partial x^{\bar{\mu}}}\frac{\partial x^\beta}{\partial x^{\bar{\nu}}}g_{\alpha\beta}$$

You can get these partial derivatives by taking derivatives of the "substitution" in the OP. This isn't a particularly clever way of getting the new metric components, but it's the most algorithmic.

For example, differentiating both sides of the first coordinate transformation with respect to $y$

$$2\cosh{R}\sinh{R}\frac{\partial R}{\partial y} = \frac{2yh^2}{b^2} \;\;\;\;\Rightarrow \;\;\;\;\frac{\partial R}{\partial y} = \frac{2yh^2}{b^2}\frac{1}{\sinh{2R}}$$

Similarly we may obtain $\partial R/\partial \theta$, $\partial \psi/\partial y$, and $\partial \psi/\partial \theta$ and construct a matrix with elements $\partial x^{\bar{\mu}}/\partial x^\alpha$. Recall

$$\frac{\partial x^\alpha}{\partial x^{\bar{\mu}}}\frac{\partial x^{\bar{\mu}}}{\partial x^\beta} = \delta^\alpha_\beta$$

So the partial derivatives of $\{R,\psi\}$ with respect to $\{y,\theta\}$ may be obtained by inverting this matrix. Careful here we're mixing two different coordinate systems when doing these calculations as a shorthand, but eventually we should convert everything to the new coordinate system.