Given $$x^2y+3y^3=7$$ I have to evaluate $y''$ at $(x,y)=(2,1)$.
I know $$y'=-\frac{2xy}{x^2+9y^2}$$ hence, at $(x,y)=(2,1)$, $$y'=-\frac{4}{13}$$
Therefore, $$y''=-\left(\frac{(2xy)'(x^2+9y^2)-(2xy)(x^2+9y^2)'}{(x^2+9y^2)^2}\right)$$ i.e., $$y''=-\frac{1}{(x^2+9y^2)^2}[(2y+2xy')(x^2+9y^2)-2xy(2x+18yy')]$$
When I substitute $x=2, y=1, y'=-4/13$, I get $$y''=-\frac{210}{2197}$$
But the answer from the book is $-210/133$. What could may be wrong?
The correct solution is what you have, note that $13^3=2197$ so this may be a possible misprint/typo