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
$$v(x)=0\Rightarrow\\
u(x)=\frac 13\ln(-3x)\\
w(x)=-2\,u(x)$$
the new metric is now $$ds^2=-(3x)^{-2/3}\,dt^2+dx^2+(3x)^{4/3}(dy^2+dz^2)$$
you obtain a metric that has the same structure as the Taub metric!!
$$ ds^2 = -x^{-2/3} dt^2 + dx^2 +x^{4/3}(dy^2 + dz^2)$$