Change of non-linear variable in derivative

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I was reading about solving elliptic equations. The example given is:

$$yu_{xx}+u_{yy}=0$$

where $y>0$.

Via the characteristic equation, the example comes to the substitutions:

$$\zeta=2y^{3/2}; \eta=3x$$

The example then says that by substitution we arrive at:

$$u_{\zeta\zeta}+u_{\eta\eta}=-\frac{u_\eta}{3\zeta}$$

But I don't see how to make this last step. I see that:

$$u_{xx}=9u_{\eta\eta}$$

However, how do we work out $u_{yy}$ in term of $\zeta$?

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Probably there is a mistake and

$$\zeta=2y^{3/2}; \eta=3x$$

Using the chain rule:

$$u_y=u_{\zeta}\zeta_y+u_{\eta}\eta_y=u_{\zeta} 3\sqrt y$$

Using the chain rule again.. (Notice that we have to use the product rule)

$$ u_{yy}= \frac{3}{2\sqrt y}u_{\zeta} + 3\sqrt y (u_{\zeta \zeta}3\sqrt y)=\frac{3}{2\sqrt y}u_{\zeta}+ 9y u_{\zeta \zeta} $$

Then:

$$yu_{xx}+u_{yy}= 9yu_{\eta \eta} + \frac{3}{2\sqrt y}u_{\zeta}+ 9y u_{\zeta \zeta} = 0 $$

So

$$u_{\eta \eta} + u_{\zeta \zeta}=\frac{-3u_{\zeta}}{18y^{\frac{3}{2}}}$$

Notice that $y^{\frac{3}{2}}= \frac{\zeta}{2}$