jmc

If $x^2+cx+9c$ is the square of a binomial, then because the coefficient of $x^2$ is $1$, the binomial must be of the form $x+a$ for some $a$. So, we have $$(x+a{)}^2=x^2+cx+9c.$$Expanding the left side, we have $$x^2+2ax+a^2=x^2+cx+9c.$$The coefficients of $x$ must agree, so $2a=c$. Also, the constant terms must agree, so $a^2=9c$, giving $c=\frac{a^2}{9}$. We have two expressions for $c$ in terms of $a$, so we set them equal to each other: $$2a=\frac{a^2}{9}.$$To solve for $a$, we subtract $2a$ from both sides: $$0=\frac{a^2}{9}-2a$$and then factor: $$0=a\left(\frac{a}{9}-2\right),$$which has solutions $a=0$ and $a=18$.

Finally, we have $c=2a$, so $c=0$ or $c=36$. But we are looking for a nonzero answer, so we can reject $c=0$. We obtain $c=\boxed{36}$.

(Checking, we find that $x^2+36x+9\cdot 36$ is indeed equal to $(x+18{)}^2$.)