Question 442912
It may be best to complete the square here:

Let's make a couple groupings here:

[x^2-6x] + [y^2 -2y] = -6

Let's complete the square with the x's.

Recall, to complete the square, take the {{{(b/2)^2}}} and make that the c term.

So our c will be (-6/2)^2 = 9

[x^2-6x + 9] + [y^2 -2y] = -6 + 9   

You must balance the equation by adding what you added to the xs.

Same process with the ys.

[x^2-6x+9] + [y^2 -2y + 1] = -6 + 9 + 1

The reason we did this was so we can now factor the xs and ys into a square.

[x^2-6x + 9] = (x-3)^2  ... it will always be {{{(x+(b/2))^2}}}

[y^2-2y + 1] = (y-1)^2

So now we have {{{(x-3)^2 + (y-1)^2 = 4}}} 

To be in standard form, we must be in this form {{{((x-h)^2)/a^2 - ((y-k)^2)/b^2 = 1}}}

We have a problem.

This is actually the equation for a circle.

{{{(x-3)^2 + (y-1)^2 = 4}}} tells us we are centered  at (3,1)  with a radius of 2. 

{{{graph(300,300,-10,10,-10,10,sqrt(4-(x-3)^2) + 1, -sqrt(4-(x-3)^2) + 1)}}}

Is it possible that either the {{{y^2}}} or {{{x^2}}} is supposed to be negative? Then you'd have yourself a hyperbola.