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

$$u(x)=\frac 23\,\ln(x)~,v(x)=-\ln(x)~,w(x)=\frac 13\,\ln(x)$$

you obtain that  the  Ricci tensor unequal zero!, so those solutions are wrong

$$\mathbf{RC}= \left[ \begin {array}{cccc} -{\frac {8}{9}}\,{x}^{4/3}&0&0&0
\\ 0&2/3\,{x}^{-2}&0&0\\ 0&0&4/9\,
{x}^{2/3}&0\\ 0&0&0&4/9\,{x}^{2/3}\end {array}
 \right] 
\ne \mathbf 0$$

**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*}
 this metric satisfy the Einstein   equation  $~\mathcal G_{\mu\nu}=0$