Multi-variable chain rule - confusion in application

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I am doing some PDEs and the wave equation that requires use of the multi-variable chain rule, which I am still getting used to.

I have $u(t, x) = v(x-ct, x+ct) = v(\varepsilon, \eta)$, where $\varepsilon = x-ct$ and $\eta = x+ct$.

Then I get the expressions for $$u_t = c \left(\frac{\partial v}{\partial \eta} - \frac{\partial v}{\partial \varepsilon}\right), u_x = \frac{\partial v}{\partial \varepsilon} + \frac{\partial v}{\partial \eta}$$

Now I am not sure how to get expressions for $u_{xx}$ and $u_{tt}$ because I am not too sure how to handle, for example: $$\frac{\partial}{\partial t} \left(\frac{\partial v}{\partial \eta}\right)$$

Help is appreciated

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With \begin{align} \eta &= x+ct\\ \varepsilon &= x-ct \end{align} We find \begin{align} 2x &= \eta+\varepsilon\\ 2ct&=\eta-\varepsilon \end{align} Thus \begin{align} \frac{\partial}{\partial \eta} &= \frac{\partial}{\partial x}\frac{\partial x}{\partial \eta}+\frac{\partial}{\partial t}\frac{\partial t}{\partial \eta}=\frac{1}{2c}\left(\frac{\partial}{\partial t}+c\frac{\partial}{\partial x} \right)\\ \frac{\partial}{\partial \varepsilon} &= \frac{\partial}{\partial x}\frac{\partial x}{\partial \varepsilon}+\frac{\partial}{\partial t}\frac{\partial t}{\partial \varepsilon}=\frac{1}{2c}\left(\frac{\partial}{\partial t}-c\frac{\partial}{\partial x} \right) \end{align}

Thus \begin{align} -4c^2 \frac{\partial^2 u}{\partial \eta \partial \varepsilon} &= \left(\frac{\partial}{\partial t}-c\frac{\partial}{\partial x}\right)\left(\frac{\partial}{\partial t}+c\frac{\partial}{\partial x}\right)u\\ &=\left(\frac{\partial^2}{\partial t^2} -c^2\frac{\partial^2}{\partial x^2}\right)u \\ &=0 \end{align}

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Note that

$$u_x = \frac{\partial v}{\partial \varepsilon}(\varepsilon, \eta) + \frac{\partial v}{\partial \eta}(\varepsilon, \eta) $$

thus in short notation

$$u_{xx}=v_{\varepsilon,\varepsilon}\varepsilon_x+v_{\varepsilon,\eta}\eta_x+v_{\eta,\varepsilon}\varepsilon_x+v_{\eta,\eta}\eta_x$$

and so on.