Is a smooth function convex near a local minimum?

No, this isn't true. Let $\phi(x)$ be a nonnegative smooth "bump" function that is zero except on $(0,1)$ where it is positive. Then $\psi(x) = \sin^2({1 \over x})\phi(x)$ is a smooth nonnegative function (define $\psi(0) = 0$) which has zeroes at $x = {1 \over k\pi}$ for positive integers $k$ but is positive between these zeroes. $\psi(x)$ will still have a local minimum at zero because $\psi(0) = 0$, but because of the humps in $\psi(x)$ between zeroes, a chord connecting $({1 \over k\pi},0)$ to $({1 \over (k +1)\pi},0)$ will lie below the graph. So $\psi(x)$ is not convex.

I should add that if $x_0$ is a local minimum of $f(x)$ such that $f^{(l)}(x_0)$ is nonzero for some $l > 0$, then it will be convex on some interval centered at $x_0$ as long as $f(x)$ is $C^{l+1}$. To see this, note that without loss of generality we may assume $l$ is minimal. By Taylor expanding one gets $$f(x) = {1 \over l!} f^{(l)}(x_0)(x - x_0)^l + O((x - x_0)^{l+1})$$ If $|x - x_0|$ is sufficiently small the remainder term will be dominated by the ${1 \over l!} f^{(l)}(x_0)(x - x_0)^l$ term. Thus the only way a local minimum can occur is if $l$ is even and $f^{(l)}(x_0) > 0$. Next note that the second derivative of $f$ has Taylor expansion given by $${d^2 f \over dx^2} = {1 \over (l-2)!} f^{(l)}(x_0)(x - x_0)^{l-2} + O((x - x_0)^{l-1})$$ The remainder term is domninated by the first term once again, which is nonnegative on an interval containing $x_0$ since $l$ is even and $f^{(l)}(x_0) > 0$. Thus the second derivative of $f$ is nonnegative on this interval and therefore the function is convex there.