Diagonalizing a constant metric tensor $g_{munu}$ at a point

Let's diagonalize $$ dtau^2 =-dt^2- 6dx,dt +17dx^2. $$

By completing the square we have $$ -dt^2- 6dx,dt +17dx^2 = - (dt +3dx)^2 +9 dx^2 +17 dx^2 = - (dt +3dx)^2 +26dx^2. $$ Choose new coordinates $T= t+3x$, $X= sqrt {26}x$.
Then $$ dtau^2= [d(x+3t)]^2-(d[sqrt{26}])^2=-dT^2 +dX^2. $$

By repeatedly completing the square you can diagonalize any quadratic form ${bf x}^TM{bf x}to (A{bf y})^TM(A{bf y})={bf y}^TD{bf y}$ where $D=A^TMA$ is diagonal. There may be many ways to do this, but as @N.Steinle says, Sylvester's law of inertia says that you will always get a diagonal metric with the same signature.

Here is an example with more variables. Consider, for example, the quadratic form $$ Q= x^2-y^2-z^2 +2xy-4xz +6yz = left(matrix{x,&!!!!y,&!!!!z}right) left(matrix{{phantom -}1&{phantom -}1&-2cr {phantom -}1&-1& {phantom -}3cr -2&{phantom -}3 & -1}right) left(matrix{xcr ycr z}right). $$ We complete the square involving $x$: $$ Q=(x+y-2z)^2 -2y^2+10yz-5z^2, $$ where the terms outside the squared group no longer involve $x$. We now complete the square in $y$:
$$ Q= (x+y-2z)^2 -(sqrt 2 y - frac 5{sqrt 2} z)^2 +frac {15}{2}z^2,$$ so that the remaining term no longer contains $y$. Thus, on setting $$ xi = x+y-2z,nonumber eta= sqrt 2 y - frac 5{sqrt 2} z,nonumber zeta = sqrt {frac{15}{2}}z,nonumber $$ we have $$ Q= xi^2 -eta^2 +zeta^2 = left(matrix{xi,&!!!!eta,&!!!!zeta}right) left(matrix{1&{phantom -}0&0cr 0&-1& 0cr 0&{phantom -}0 & 1}right) left(matrix{xicr etacr zeta}right). $$