A convex $n$-gon can be cut into $n-2$ triangles by just choosing a vertex and drawing all the diagonals from that vertex, so $3n$ obtuse triangles can be reduced to $3n-6$.
We can do a bit better. E.g., if a convex quadrangle is not a rectangle, then it has at least one obtuse angle, so we can cut off an obtuse triangle incorporating that angle, and just need three more triangles to finish the job, four triangles in all. Every convex $n$-gon, $n\ge5$, has one or more obtuse angles, which we can use to cut off triangles, to reduce the $3n-6$ further.
EDIT ––– Taking this observation to its logical conclusion, we can see that any convex $n$-gon, other than a rectangle, can be partitioned into $n$ obtuse triangles (a rectangle can be partitioned into six obtuse triangles).
This is certainly true for $n=3$. Any convex $n$-gon with $n\ge4$, except the rectangle, has at least one obtuse angle; cutting off the triangle containing this obtuse angle and the two adjacent vertices yields a convex $n-1$-gon, so induction yields the claimed result, provided that when going from a pentagon to a quadrilateral we can avoid forming a rectangle. But if a pentagon is an obtuse triangle glued to a rectangle, then the pentagon has two other obtuse angles in addition to the one we used, and using either one of these other obtuse angles, we get a quadrilateral that can't be a rectangle.
And, we're done.