I stumbled upon Fermat's theorem while reading Steig Larsson's "The Girl Who Played With Fire":
$x^3 = y^3 + z^3 \tag{Eq. 1}$
As far as it goes, it is an academic tweak of the Pythagorean theorem.
$x^2 = y^2 + z^2\tag{Eq. 2}$
So differentiating Eq. $1$ with respect to $x$, we get
$$ x^2 = (y^2)\frac{\mathrm dy}{\mathrm dx} + (z^2)\frac{\mathrm dz}{\mathrm dx}\tag{Eq. 3}$$
Now LHS of Eq. $2$ and $3$ are equal.
Eq. $3$ has scalars of $\mathrm dy/\mathrm dx$ and $\mathrm dz/\mathrm dx$ which are slopes.
Now for the RHS of Eq. $2$ and $3$ to be equal, we would want that both $y$ slope and $z$ slope be unity or $1$.
But if two sides have the same slope of $1$, then they are parallel.
But it is a given that in a right angled triangle or for that sake, any triangle cannot have parallel lines. So it is impossible for RHS of Eq. $2$ and $3$ to be same.
Does this provide a proof for Fermat's claim?
P.S Bare my follies if my question turns out to be a very stupid one. Was just curious.
There are several mistakes here.
First, let's look back at the Pythagorean Theorem. This does not say "$x^2=y^2+z^2$," since that's blatantly false in general; rather, it says that if $x, y, z$ are the sides of a right triangle with $x$ being the hypoteneuse, then $x^2=y^2+z^2$.
A related fact is that there are infinitely many triples of real numbers $(x, y, z)$ satisfying $x^2=y^2+z^2$. Of course, this is trivial: take any $x>y>0$, and let $z=\sqrt{x^2-y^2}$.
Slightly more surprising is that there are infinitely many triples of integers satisfying that equation: e.g.
$5^2=3^2+4^2$
$13^2=12^2+5^2$
And infinitely many others. (See e.g. here.)
Now let's look at Fermat's Last Theorem. FLT is the statement that the fact above fails for powers $>2$: namely, it says
You mention specifically the case $c=3$; note that this is just a special case of FLT, and was proved long before FLT.
Now the first problem with your argument essentially comes down to this: even if we had $x^3=y^3+z^3$, there is no reason to believe that $(x, y, z)$ is also a Pythagorean triple! So you can't hope to prove even the $c=3$ case of FLT by anything like what you write.
Moreover, since we're looking only at integer solutions (note that there trivially are real number solutions to $x^3=y^3+z^3$ - again, pick $x>y>0$ and let $z=\sqrt[3]{x^3-y^3}$), this is fundamentally a discrete problem - calculus isn't really going to come into play.
EDIT: Another, maybe more concrete, mistake occurs when you write
If $a+b=ca+db$, it is true that $c=d=1$ is one way for this to happen. But it's not the only way! For instance, $$2+2={1\over 2}\cdot 2+{3\over 2}\cdot 2.$$ So right there, you cannot deduce the thing you want (which, as I said above, isn't even relevant to the problem).