This particular problem can dealt with algebraically by completing squares, equating coefficients and subtracting constants. (I am making the assumptions this is a left hand side whose right hand side is 0).
pol = -0.433284 - 0.758719 x + 0.00289158 x^2 - 0.443672 y +
0.00149027 y^2;
cr = CoefficientRules[pol, {x, y}]
{p, q, r, s, t} = cr[[All, 2]];
{h, k} = 0.5 {-q/p, -s/r}
cons = p h^2 + r k^2
This yields for {h,k}:=
{131.195, 148.856}
The rearranged polynomial:
trp[x_, y_] := p (x - h)^2 + r (y - k)^2 - cons + t

and note:
Expand[trp[x, y]] == pol
yields True
.
If the aim is to get to $\frac{(x-h)^2}{a^2}+\frac{(y-k)^2}{b^2}=1$
{a, b} = Sqrt[
1/Coefficient[(trp[x + h, y + k] + cons - t)/(cons - t), {x^2,
y^2}]]
This yields:
{169.652, 236.316}
The boolean is hampered by numericals but to show the rearrangement is equivalent:
stf[x_, y_] := (x - h)^2/a^2 + (y - k)^2/b^2 - 1
Column[{TraditionalForm[pol],
TraditionalForm[Expand[(cons - t) stf[x, y]]]}]

Putting it all together:
TraditionalForm[
StringForm[
"\!\(\*FractionBox[SuperscriptBox[\((x - `1`)\), \(2\)], \
SuperscriptBox[\(`2`\), \
\(2\)]]\)+\!\(\*FractionBox[SuperscriptBox[\((y - `3`)\), \(2\)], \
SuperscriptBox[\(`4`\), \(2\)]]\)=1", h, a, k, b]]

Visualizing:
Grid[{{TraditionalForm[pol == 0],
TraditionalForm[stf[x, y] == 0]}, {ContourPlot[
pol == 0, {x, -500, 500}, {y, -500, 500}, ImageSize -> 300],
ContourPlot[stf[x, y] == 0, {x, -500, 500}, {y, -500, 500},
ImageSize -> 300]}}, Frame -> All]
