= Solution
There is a global-domain issue in the printed question. Write $D=a^2x^2+b^2y^2>0$ away from the origin. Direct <partial differentiation> gives
$$
P_y=-\frac{ab(a^2x^2-b^2y^2)}{D^2}
=Q_x.
$$
Thus the <differential one-form> is a <closed differential form> for \b[every real $a,b\ne0$]. The <Poincaré lemma> makes it locally an <exact differential form> on small neighborhoods avoiding the origin. There is no additional condition such as $a^2=b^2$.
However, the stated domain is the whole <punctured plane>, which is not <simply connected>. To test global exactness, use the closed <ellipse> $x=\cos t/a$, $y=\sin t/b$, $0\leq t\leq2\pi$. Here $D=1$ and
$$
P\,dx+Q\,dy=ab(x\,dy-y\,dx)=dt,\qquad
\oint(P\,dx+Q\,dy)=2\pi.
$$
The <line integral> of a globally <exact differential form> around any closed curve would be zero. Consequently
$$
\boxed{\text{globally exact on }\mathbb R^2\setminus\{0\}:\ \text{no allowed }a,b;\qquad
\text{locally exact}:\ \text{all }a,b\ne0.}
$$
The distinction is precisely that of the <elliptically scaled angular one-form>. The remaining parts can be answered consistently using a local angle branch, as follows.
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