Official solution
Equation of ellipse,
\[{{{x^2}} \over {{a^2}}} + {{{y^2}} \over {{b^2}}} = 1\]
As ellipse passes through (0, 3)
\[\therefore\,\,\,\] \[{{{0^2}} \over {{a^2}}} + {{{3^2}} \over {{b^2}}} = 1\]
\[ \Rightarrow \] b2 = 9
\[\therefore\,\,\,\] Equation of ellipse becomes,
\[{{{x^2}} \over {{a^2}}} + {{{y^2}} \over 9} = 1\]
Differentiating w.r.t x, we get,
\[{{2x} \over {a{}^2}}\] + \[{{2y} \over 9}\] . \[{{dy} \over {dx}} = 0\]
\[ \Rightarrow \] \[{x \over {{a^2}}}\] = - \[{y \over 9}.{{dy} \over {da}}\]
\[ \Rightarrow \] \[{x \over {{a^2}}} = - {y \over 9}.y'......\] (1)
We got earlier,
\[{{{x^2}} \over {{a^2}}} + {{{y^2}} \over 9}\] = 1
\[ \Rightarrow \] \[{x \over {{a^2}}}.x + {{{y^2}} \over 9} = 1\]
putting value of equation (1) here,
\[ {-{y\,y'} \over 9}.x + {{{y^2}} \over 9} = 1\]
\[ \Rightarrow \] - xyy' + y2 = 9
\[ \Rightarrow \] xyy' - y2 + 9 = 0