I follow the web documentation
I) Metric Ansatz
$$\mathbf G= \left[ \begin {array}{cccc} -{{\rm e}^{2\,u \left( x \right) }}&0&0&0 \\ 0&{{\rm e}^{2\,v \left( x \right) }}&0&0 \\ 0&0&{{\rm e}^{2\,w \left( x \right) }}&0 \\ 0&0&0&{{\rm e}^{2\,w \left( x \right) }} \end {array} \right] $$
II) Ricci Tensor $~\mathbf{RC}~$
$$RC_{1,1}={\frac {d^{2}}{d{x}^{2}}}u \left( x \right) + \left( {\frac {d}{dx}}u \left( x \right) \right) ^{2}- \left( {\frac {d}{dx}}u \left( x \right) \right) {\frac {d}{dx}}v \left( x \right) +2\, \left( { \frac {d}{dx}}u \left( x \right) \right) {\frac {d}{dx}}w \left( x \right) =0\tag 1$$
$$ RC_{2,2}=\left( {\frac {d}{dx}}u \left( x \right) \right) {\frac {d}{dx}}w \left( x \right) +{\frac {d^{2}}{d{x}^{2}}}w \left( x \right) +2\, \left( {\frac {d}{dx}}w \left( x \right) \right) ^{2}- \left( { \frac {d}{dx}}v \left( x \right) \right) {\frac {d}{dx}}w \left( x \right) =0\tag 2$$
$$RC_{3,3}={\frac {d^{2}}{d{x}^{2}}}u \left( x \right) + \left( {\frac {d}{dx}}u \left( x \right) \right) ^{2}- \left( {\frac {d}{dx}}u \left( x \right) \right) {\frac {d}{dx}}v \left( x \right) +2\,{\frac {d^{2}} {d{x}^{2}}}w \left( x \right) +2\, \left( {\frac {d}{dx}}w \left( x \right) \right) ^{2}-2\, \left( {\frac {d}{dx}}v \left( x \right) \right) {\frac {d}{dx}}w \left( x \right) =0\tag 3$$
now if you substitute the function that the author obtains
$$u(x)=\frac 23\,\ln(x)~,v(x)=-\ln(x)~,w(x)=\frac 13\,\ln(x)$$
the Ricci tensor
$$\mathbf{RC}= \left[ \begin {array}{cccc} -{\frac {8}{9}}\,{x}^{4/3}&0&0&0 \\ 0&2/3\,{x}^{-2}&0&0\\ 0&0&4/9\, {x}^{2/3}&0\\ 0&0&0&4/9\,{x}^{2/3}\end {array} \right] \ne \mathbf 0$$
III the Solution
solving the equations (1),(2) and (3) you obtain
$$u(x)=\text{arbitrary}\\ v(x)=-3\,u(x)+ln(u'(x))\\ w(x)=-2\,u(x)$$
with
$$u(x)=\frac 23\ln(x)\quad\Rightarrow\\
v(x)=-2\ln(x)+\ln\left(\frac{2}{3x}\right)\\
w(x)=-\frac 43\ln(x)$$
the new metric is now
$$\mathbf G=\left[ \begin {array}{cccc} -{x}^{4/3}&0&0&0\\ 0&{ {\rm e}^{-4\,\ln \left( x \right) +2\,\ln \left( 2/3\,{x}^{-1} \right) }}&0&0\\ 0&0&{x}^{-8/3}&0 \\ 0&0&0&{x}^{-8/3}\end {array} \right] $$
with $$X=~{{\rm e}^{-4\,\ln \left( x \right) +2\,\ln \left( 2/3\,{x}^{-1} \right) }}\quad\Rightarrow\\ x=\frac 13\,\sqrt [3]{2}{3}^{2/3}\sqrt [6]{{X}^{-1}}\\ dx=-\frac{1}{18}\,{\frac {\sqrt [3]{2}{3}^{2/3}{\it dX}}{ \left( {X}^{-1} \right) ^{5/6}{X}^{2}}}\quad\Rightarrow $$
\begin{align*} &G_{11}=-\frac{1}{3}\,\frac{2^{4/9}\,3^{5/9}}{X\,\left(\frac{1}{X}\right)^{7/9}}\\ &G_{22}=\frac{1}{108}\,\frac{2^{2/3}\,3^{1/2}}{X^2\,\left(\frac{1}{X}\right)^{2/3}}\\ &G_{33}=\frac{1}{2}\,\frac{3^{8/9}\,2^{1/9}}{\left(\frac{1}{X}\right)^{4/9}}\\ &G_{44}=G_{33} \end{align*} i