Crazy diffy!

Putting $p = \frac{dx}{dy}$ , the equation becomes $\frac{dp}{dy} \; =\; \tfrac13\sqrt{1 + p^2} \;,$ which has solution $p \; =\; \sinh\big(\tfrac13y+c) \;.$ Thus $\frac{dx}{dy} \; = \; \sinh\big(\tfrac13y+c) \;,$ and so $x \; = \; 3\cosh\big(\tfrac13y+c\big) + d \;.$ Since $y = 0$ when $x=3$ we have $3 = 3\cosh c + d$ , and hence $x \; = \; 3\cosh\big(\tfrac13y+c\big) + 3 - 3\cosh c \;.$ Since $p = \frac{dx}{dy} = 0$ when $x=0$ , we see that $y = -3c$ at this point, and so $0 \; = \; x \; = \; 3\cosh 0 + 3 - 3\cosh c \; = \; 6 - 3\cosh c$ so that $\cosh c = 2$ , and hence $x \; = \; 3\cosh\big(\tfrac13y + \cosh^{-1}2\big) - 3 \;.$ When $y=6$ we have $x = 3\cosh\big(2 + \cosh^{-1}2\big) - 3 \,=\, 38.418893464\,$ making the answer $\boxed{38}$ .