In case of Folium of Descartes $x^3+y^3=3axy$ , prove that the radius of curvature at the point $(3a/2,3a/2)$ is numerically equal to $3\sqrt2 a/16$.
The solution given in the book is as follows:
If $f(x,y)=x^3+y^3=3axy=0$ we may easily obtain $f_x(3a/2,3a/2)=(9/4)a^2=f_y(3a/2,3a/2);f_{xx}(3a/2,3a/2)=9a=f_{yy}(3a/2,3a/2);f_{xy}(3a/2,3a/2)=-3a$. Now, we know that radius of curvature at a point on a curve (for $f(x,y)=0$) $\rho=\frac{(f_x^2+f_y^2)^{\frac{3}{2}}}{f_{xx}f_y^2-2f_{yx}f_xf_y+f_{yy}f_x^2}$ and hence substituting the values we get $\rho=3\sqrt2 a/16$.
However, I am not quite getting it. I have a really fundamental doubt . What do they mean by $f_x, f_{xx},f_{yy},f_{xy}$, etc ? I have no idea about it.