You fell in the trap. A relative change can be computed wrt the old value or the new value, you need to be specific about that.
$$\frac{371-350}{350}=\frac{371}{350}-1=6.00\%$$
$$\frac{371-350}{371}=1-\frac{350}{371}=5.66\%$$
For a similar reason,
$$\begin{align}100+10\%&=110,\\110-10\%&=99\ !?!\end{align}$$
Neither of these is correct, although they are both approximately correct if the errors are small. The exact answer is the following. If you're multiplying a quantity which you are uncertain about by a factor of $1 \pm p$ with a quantity which you are uncertain about by a factor of $1 \pm q$, then you are uncertain about their product by a factor of
$$(1 \pm p)(1 \pm q) = 1 \pm p \pm q + pq.$$
This is simple arithmetic. For example, if you have 10% uncertainty about the first quantity and 8% uncertainty about the second quantity, then you're uncertain about their product by a factor of
$$(1 \pm 0.1)(1 \pm 0.08) = 1 \pm 0.18 + 0.008.$$
Note that this is not symmetric about $1$; the upper bound is $1.1 \times 1.08 = 1.188$ and the lower bound is $0.9 \times 0.92 = 0.828$. You can say conservatively that the uncertainty in the product is $1 \pm 0.188$, or 18.8%, but this loses a little bit on the lower end.
If $p$ and $q$ are small then $pq$ is very small and $\pm p \pm q + pq$ is approximately $\pm p \pm q$, which is where "add the relative errors" comes from, but it's worth knowing that this is an approximation that breaks down if $p$ and $q$ are not small (or if you are multiplying many terms).
If you read the second link more carefully it tells you to compute the square-root-of-sum-of-squares for the absolute error of a sum (at least I think that's what it's doing; it's not very clear since the term "SE" hasn't been defined at all); this isn't a method being suggested for the relative error of a product at all. The main reason you'd compute the error this way is if you have reason to believe that your errors are well-modeled by independent normal / Gaussian distributions. This square-root-of-sum-of-squares behavior governs how independent Gaussians add: we have
$$N(0, \sigma_1) + N(0, \sigma_2) \sim N(0, \sqrt{\sigma_1^2 + \sigma_2^2}).$$
But this is a specific modeling assumption that may break down. The exact expression I give above gives worst-case bounds which are independent of the nature of the error as long as you actually know a bound on it.
Best Answer
It depends on what you are trying to do.
If you start out with $6$ and then it increases to $9$, and you want to know by how much the original value increased (or decreased), then you use **[(new value - old value)/old value] * 100. So, here, it would be $\dfrac{9 - 6}{6}*100 = \dfrac{3}{6}*100 = 50$%. (If when using this formula, your percent is a negative number, then that is the percentage it decreased by.