Unless the closed form solution is extremely expensive to compute, it generally is the way to go when it is available. However,
For most nonlinear regression problems there is no closed form solution.
Even in linear regression (one of the few cases where a closed form solution is available), it may be impractical to use the formula. The following example shows one way in which this can happen.
For linear regression on a model of the form $y=X\beta$, where $X$ is a matrix with full column rank, the least squares solution,
$\hat{\beta} = \arg \min \| X \beta -y \|_{2}$
is given by
$\hat{\beta}=(X^{T}X)^{-1}X^{T}y$
Now, imagine that $X$ is a very large but sparse matrix. e.g. $X$ might have 100,000 columns and 1,000,000 rows, but only 0.001% of the entries in $X$ are nonzero. There are specialized data structures for storing only the nonzero entries of such sparse matrices.
Also imagine that we're unlucky, and $X^{T}X$ is a fairly dense matrix with a much higher percentage of nonzero entries. Storing a dense 100,000 by 100,000 element $X^{T}X$ matrix would then require $1 \times 10^{10}$ floating point numbers (at 8 bytes per number, this comes to 80 gigabytes.) This would be impractical to store on anything but a supercomputer. Furthermore, the inverse of this matrix (or more commonly a Cholesky factor) would also tend to have mostly nonzero entries.
However, there are iterative methods for solving the least squares problem that require no more storage than $X$, $y$, and $\hat{\beta}$ and never explicitly form the matrix product $X^{T}X$.
In this situation, using an iterative method is much more computationally efficient than using the closed form solution to the least squares problem.
This example might seem absurdly large. However, large sparse least squares problems of this size are routinely solved by iterative methods on desktop computers in seismic tomography research.
Your question is interesting, because it is a starting point into optimization in general.
It is maybe better to start with a concrete example on the existence of analytical solution. If you look at your standard polynomial of 2nd degree, $f(x)=ax^2+bx+c$. You have a formula for the zeroes of the function, which is an analytical solution. If you have a polynomial of order 5 or higher, no such formula exists.
This is of course not an optimization problem as in minimizing or maximizing a function, but when you find where the derivative is zero, you are essentially looking for a zero of a function, namely the derivative.
So it seems that you cannot always solve a problem with pen and paper. This is also dependent on the scale of the problem. Sometimes you need to estimate billions of parameters, and it is just not feasible for a human to work out an analytical solution or even find out if it exists.
In the case of ordinary least squares, or fitting a linear model, an analytical solution exists. But it is very easy to change the model slightly, such that no such solution exists, (at least not one we know of). An example of this is the lasso.
So the use of numerical methods is completely justified, both by the fact that an analytical solution may not exist, or it is not feasible to work out such a solution.
Best Answer
Towards Data Science isn't a reliable website, and the text you've quoted is, unfortunately, nonsense.
What they meant to say, I hope, is that "analytical problems are Determinstic [...]", etc.
I won't explain the difference between analytic and numeric approaches here, because there are lots of good sources, but going by this paragraph I'm going to say the post you read isn't one of them.
EDIT: OK, I'll explain a bit
Part of the problem is that there are a lot of partially overlapping terms. Very roughly speaking, you have:
There are plenty of other ways to slice this up, but these should be plenty to get started!