= Solution
Use the real <eigenfunctions>, or insert a complex conjugate on the first factor for complex functions. Multiply $-(py_n')'=n^2wy_n$ by $y_m$, do the same with $m,n$ interchanged, and subtract. <Integration by parts> gives
$$
(n^2-m^2)\int_{-1}^1wy_ny_m\,dx=\left[p(y_ny_m'-y_my_n')\right]_{-1}^1=0.
$$
The boundary term vanishes because both functions and derivatives are bounded while $p\to0$. For distinct squared indices, <orthogonality of Sturm-Liouville eigenfunctions> therefore yields
$$
\boxed{\int_{-1}^1\frac{y_n(x)y_m(x)}{\sqrt{1-x^2}}\,dx=0\qquad(n^2\ne m^2).}
$$
A second integration by parts, using the equation for $y_n$, gives
$$
\int_{-1}^1p\,y_n'y_m'\,dx=[py_n'y_m]_{-1}^1-\int_{-1}^1y_m(py_n')'dx=n^2\int_{-1}^1wy_ny_m\,dx.
$$
Thus <derivative orthogonality when the Sturm-Liouville potential vanishes> gives
$$
\boxed{\int_{-1}^1y_n'y_m'\sqrt{1-x^2}\,dx=0\qquad(n^2\ne m^2).}
$$
\b[The index qualification matters.] The source initially allows integer indices, so $n\ne m$ alone does not imply distinct eigenvalues: $m=-n$ gives the same $n^2$. Taking the conventional nonnegative labels $n,m\geq0$ makes the stated conclusion valid. For the polynomial solutions below, $y_{-n}=y_n$, so opposite indices are an explicit counterexample to an unqualified orthogonality claim.
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