Very good intuition. The only flaw in your thinking is that in the original equation,
$\frac{dy}{dx}$ is multiplied by $x$.
Therefore, when $x = 0$, it becomes irrelevant what $\frac{dy}{dx}$ is.
Addendum
Reaction to OP's comment/reaction re 10-3-2020.
The original equation is $x\frac{dy}{dx} - y = x^3.$
You discovered that the solution is not one equation but a family
of equations, represented by
$y = \frac{x^3}{2} + cx ~\Rightarrow ~\frac{dy}{dx} = f'(x) = \frac{3x^2}{2} + c.$
Your intuition then rebelled, intuiting (in effect):
Something is wrong here.
Consider two separate solutions:
$f_1(x) = \frac{x^3}{2} + c_1x.$
$f_2(x) = \frac{x^3}{2} + c_2x ~: ~c_2 \neq c_1$
As the value of $c$ changes, so does the value of $\frac{dy}{dx}.$
This means, that at any given value of $x$,
$f'_1(x)$ and $f'_2(x)$ will be unequal.
How can a family of functions, as represented by $f_1(x)$ and $f_2(x)$
each of whom must have a different value for $f'(x_0)$ at one
specific value of $x_0$ ever intersect?
Answer:
My original answer gave only a mathematical explanation for why
(for example) $f_1(x)$ and $f_2(x)$ can intersect at the origin
despite the fact that $f'_1(0) \neq f'_2(0).$
Intuitively, consider the alternative set of two differential equations:
$f''(x) = 0,$ for all $x~~$ combined with $~~f(x) = 0.$
The above two equations will be satisfied by the family of equations:
$f(x) = cx,~$ all of which intersect at $x=0$.
Just because each member of the family has a different derivative at $x=0$
doesn't mean that they can't all intersect at $x=0$.
As far as the specific terminology in your comment, although I have had
a brief exposure to the concepts of isocline and directional field, I
lack the professional experience to grapple with these concepts and explain
the supposed flaw in your intuition (if any).
It seems to me that you are asking a legitimate question that is simply
beyond my expertise. If I were you, and I couldn't come to terms with
your pending question
in 24 hours, and if no one else responded because mathSE queries tend to
get lost in the shuffle, then I would:
(1)
Create a new mathSE query. In that new query, provide a link to this query.
Indicate that the new query is a followup to this query, re there is still
an unresolved issue. Spell out the issue as clearly as you can, being
sure to use very similar syntax as in your comment.
This way you will (in effect) be (quite reasonably) attempting to force
qualified mathSE reviewers to focus on the issue that your perspective
deems as pending.
(2)
In this query, leave your comment just where it is. However, add an
addendum to your original query in this posting. In this addendum,
repeat the pending question of your pertinent comment. Indicate that
from your perspective, this is a pending question. Indicate (also in
the addendum) that you are construing the pending question to be a
second question, and you have therefore initiated a 2nd mathSE query.
Provide a link to the 2nd mathSE query in the addendum of this original query.
Best Answer
Ignoring the $f=0$ solution, we can rearrange the equation to see that
$$\frac{f'}{\sqrt{f}} = 2(\sqrt{f})' = \frac{1}{2x}\sqrt{f}-1$$
a linear equation in $\sqrt{f}$, which we can solve by integrating factor
$$\int\left(\frac{\sqrt{f(x)}}{x^{\frac{1}{4}}}\right)' = \int-\frac{1}{2x^{\frac{1}{4}}} \implies \sqrt{f(x)} = Cx^{\frac{1}{4}}-\frac{2}{3}x$$
However, at this stage, $C$ is not allowed to be any real number - it is dependent on the domain of the function. For positive $x$ we have that
$$Cx^{\frac{1}{4}}-\frac{2}{3}x\geq 0 \implies C \geq \frac{2}{3}x^{\frac{3}{4}}$$
Your choice of $C=1$ was invalid because extending the domain of the function to $20$ puts a lower bound on $C$ to be
$$C \geq \frac{2}{3}(20)^{\frac{3}{4}} \approx 6.3049$$
Alternatively, once $C$ is chosen, it provides an upper bound to the domain of the function, in which case $C=1$ puts the maximal extent of the interval of existence to be
$$0 < x \leq \left(\frac{3}{2}C\right)^{\frac{4}{3}} = \left(\frac{3}{2}\right)^{\frac{4}{3}} \approx 1.717$$
You cannot have arbitrarily large $x$.