Let $X$ be a smooth degree $d$ ($d >5$) surface in $\mathbb{P}^3$. Let $\pi:\tilde{X} \to X$ be a blow-up of $X$ at a point. When is it possible to embed $\tilde{X}$ into $\mathbb{P}^3$?
In general, when can we embed a projective surface in $\mathbb{P}^3$? When is the resulting surface smooth?
Let $Y\subset \mathbb{P}^3$ be a smooth projective surface of degree $l$. If $l\geq 4$, then $K_Y=(l-4)H|_Y$ is nef (really either $0$ or very ample), and hence $Y$ cannot contain a $(-1)$ curve. If $l\leq 2$, then $Y$ is either a plane or a quadric, neither of which contain $(-1)$ curves. This leaves $l=3$. The cubic indeed contains $(-1)$ curves, but the surface obtained by blowing down one of those cannot be embedded into $\mathbb P^3$ since the cubic surface is rational and hence so is its blow down, so if it is embedded into $\mathbb P^3$ it would have to have degree at most $3$. Comparing Picard numbers (for instance, or pretty much anything else you can think of) you can see that this blow down is not a plane, a quadric or a cubic. However, you don't even need any of that since your $X$ is of general type and hence cannot be birational to a cubic (which is the only surface in $\mathbb P^3$ that contains a $(-1)$ curve.)
The blow-up of this surface will contain the exceptional curve s.t. its intesection with the canonical class of the blown-up surface is $-1$; sind canonical class of a smooth surface of degree $m$ in $\mathbb P^3$ is $(m-4)$ times hyperplane section, this is possible only if the blown-up surface is embedded as a surface of degree $\le 3$. Plane and quadric contain no exceptional curves; a smooth cubic with one blown down exceptional curve is isomorphic to the intersection of two quadrics, which cannot be embedded in $\mathbb P^3$. So the answer to your question seems to be 'never'.
