The axes of the new system Ox'y'z' are the bisectors of the…
The axes of the new system $Ox'y'z'$ are the bisectors of the angles $\angle xOy$, $\angle yOz$, $\angle zOx$ of the rectangular system $Oxyz$. Find the transformation formulas (with the same unit of length on all axes).
Explanation of Method
New axes along the bisectors of the old coordinate angles form an oblique system.
Step 1: New unit vectors
\[ \vec e_{x'}=\tfrac1{\sqrt2}(1,1,0),\qquad \vec e_{y'}=\tfrac1{\sqrt2}(0,1,1),\qquad \vec e_{z'}=\tfrac1{\sqrt2}(1,0,1) \]
Each pairwise dot product is $\tfrac12$, so every pair of new axes makes an angle of $60^\circ$. The new system is oblique, and the transformation is affine, not orthogonal.
Step 2: Direct formulas
From $\vec r=x'\vec e_{x'}+y'\vec e_{y'}+z'\vec e_{z'}$: \[x=\frac{x'+z'}{\sqrt2},\qquad y=\frac{x'+y'}{\sqrt2},\qquad z=\frac{y'+z'}{\sqrt2}\]
Step 3: Inverse formulas
With $N=\begin{pmatrix}1&0&1\\1&1&0\\0&1&1\end{pmatrix}$: $\det N=2$ and $N^{-1}=\tfrac12\begin{pmatrix}1&1&-1\\-1&1&1\\1&-1&1\end{pmatrix}$. The transformation matrix is $A=\tfrac1{\sqrt2}N$, so
\[ A^{-1}=\sqrt2\,N^{-1}=\frac{\sqrt2}{2}\begin{pmatrix}1&1&-1\\-1&1&1\\1&-1&1\end{pmatrix}. \]
\[x'=\frac{x+y-z}{\sqrt2},\qquad y'=\frac{-x+y+z}{\sqrt2},\qquad z'=\frac{x-y+z}{\sqrt2}\]
Step 4: Verification
The point $\vec e_{x'}=\left(\tfrac1{\sqrt2},\tfrac1{\sqrt2},0\right)$ must have $x'=1,\ y'=z'=0$: \[x'=\frac{\frac1{\sqrt2}+\frac1{\sqrt2}}{\sqrt2}=1,\qquad y'=\frac{-\frac1{\sqrt2}+\frac1{\sqrt2}}{\sqrt2}=0,\qquad z'=0.\ \checkmark\]
Substituting the direct formulas: $\tfrac{x+y-z}{\sqrt2}=\tfrac{(x'+z')+(x'+y')-(y'+z')}{2}=x'$. $\checkmark$
$\det A=\tfrac{2}{2\sqrt2}=\tfrac1{\sqrt2}\ne0$. This is the volume of the parallelepiped on $\vec e_{x'},\vec e_{y'},\vec e_{z'}$.
Result
\[\boxed{\,x=\frac{x'+z'}{\sqrt2},\qquad y=\frac{x'+y'}{\sqrt2},\qquad z=\frac{y'+z'}{\sqrt2}\,}\]
\[\boxed{\,x'=\frac{x+y-z}{\sqrt2},\qquad y'=\frac{-x+y+z}{\sqrt2},\qquad z'=\frac{x-y+z}{\sqrt2}\,}\]
Remark
Correction: the source inverts $A=\tfrac1{\sqrt2}N$ as $\tfrac{\sqrt2}{2}\cdot N^{-1}$ and gets the factor $\tfrac{\sqrt2}{4}$. The correct factor is $\sqrt2\cdot\tfrac12=\tfrac{\sqrt2}{2}=\tfrac1{\sqrt2}$. The factor $\tfrac{\sqrt2}4$ halves all coordinates: $\vec e_{x'}$ gets $x'=\tfrac12$ instead of $1$. The “oblique sense” explanation offered in the source is not valid, since oblique coordinates of $\vec e_{x'}$ are exactly $(1,0,0)$. The source's consistency check also contains the slip $\tfrac{\sqrt2}{4}\cdot\tfrac{2x'}{\sqrt2}=x'$; the left side equals $\tfrac{x'}{2}$.