Polar to Cartesian relations:
x=Rsintheta
y=R cos theta
Our equation:
3x^2+4y^2-6x=9
Substitute:
3R^2sin^2theta+4R^2cos^2theta-6Rsintheta=9
Simplify:
3R^2(sin^2theta+cos^2theta)+R^2cos^2theta-6Rsintheta=9
3R^2(1+cos^2theta)-6Rsintheta=9
R^2(1+cos^2theta)-2Rsintheta-3=0
We can if we wish rewrite this in the form R=f(theta) by solving the quadratic here in R. As the original equation wasn't given in the form y=g(x) we shouldn't feel obliged to, but it might be a useful formulation.
R=(2sintheta+-sqrt(4sin^2theta+12(1+cos^2theta)))/(2(1+cos^2theta))
R=(sintheta+-sqrt(sin^2theta+3(1+cos^2theta)))/(1+cos^2theta)
R=(sintheta+-sqrt(4+2cos^2theta))/(1+cos^2theta)