We have given

`y^2+12x=2y-13`

shift the variables and we have

`y^2-2y=-12x-13`

make lhs perfect square so ,we can write

`y^2-2y+1=-12x-13+1`

`(y-1)^2=-12x-12`

Which is equation of parabola and defined if

`-12x-12>=0`

`-12x>=12`

`x<=-1`

parabola is open left i.e second and third quadrant .

`(y-1)^2=-12(x+1)`

`vertex` is (-1,1)

What have we to do with?

`y^2-2y+13=12x`

`(y-1)^2+12=12x`

`(y-1)^2=12(x-1)`

`y-1=2sqrt(3(x-1))`

`y=1+2sqrt(3(x-1))`

`x=0 ` not for real value of `y`

`y=1+-2i sqrt(3)`

`y=0` `x=1`

explicit function `y=y(x)` defined for `x>=1`

We note function hasn't critical points, indeed:

`y'(x)=1/sqrt(3(x-1))`

that never will be like 0.

Never if we concern about original function:

`f(x,y)=y^2-2y-12x+13`

we wl lfaound critcal points:

Indeed: `(del f)/(del y) =2(y-1)` zero for `y=1`

but `(del f)/(del x)=-12` that won'tever be zero.