Proving $F(x,y)=\left(x,y,\int_x^yf(z^2)dz\right)$ is differentiable and computing $\nabla F(x,y)$ 
Let $f:\Bbb R\to\Bbb R$ be a function of the class $C^1$ and let $F:\Bbb R^2\to\Bbb R^3$ be given by $$F(x,y)=\left(x,y,\int_x^yf(z^2)dz\right).$$
Prove $F$ is differentiable and compute $\nabla F(x,y).$

My thoughts:
First, $F$ is differentiable if and only if all of its components are, so it remains to show $F_3(x,y)=\int_x^yf(z^2)dz$ is differentiable.
I think I should use the following results:
$\underline{\boldsymbol{\text{theorem } 11.1:}}$

Let $f:A=[a,b]\times[c,d]\to\Bbb R$ be continuous s. t. $\frac{\partial f}{\partial y}$ exists and is continuous on $A.$ Suppose $F:[c,d]\to\Bbb R$ is given by $$F(y)=\int_a^bf(x,y)dx.$$ Then $F$ is differentiable on $[c,d]$ and $F'(y)=\int_a^b\frac{\partial f}{\partial y}(x,y)dx.$

And its corollary:
$\underline{\boldsymbol{\text{corollary }11.2:}}$

Let $f:A=[a,b]\times[c,d]\to\Bbb R$ is continuous s.t. $\frac{\partial f}{\partial y}$ exists and is continuous on $A.$ Let $u,v:[c,d]\to[a,b]$ be of class $C^1$ and suppose $F:[c,d]\to\Bbb R$ is given by $$F(y)=\int_{u(y)}^{v(y)}f(x,y)dx.$$ Then, $F$ is differentiable on $[c,d]$ and $$F'(y)=f(v(y),y)v'(y)-f(u(y),y)u'(y)+\int_{u(y)}^{v(y)}\frac{\partial f}{\partial y}(x,y)dx.$$

Both results have been proven and I got some additional insight into the corollary here.
Let $G, H:\Bbb R\to\Bbb R, G(t):=F_3(0,t), H(t):=F_3(t,0).$
We could write: $$\begin{aligned}F_3(x,y)&=\int_x^yf(z^2)dz\\&=\int_0^yf(z^2)dz-\int_0^xf(z^2)\\&=F_3(0,y)-F(0,x)\\&=G(y)-G(x)\end{aligned}$$ and $$\begin{aligned}F_3(x,y)&=\int_x^yf(z^2)dz\\&=\int_x^0f(z^2)+\int_0^yf(z^2)dz\\&=\int_x^0f(z^2)dz-\int_y^0f(z^2)dz\\&=F_3(x,0)-F_3(y,0)\\&=H(x)-H(y)\end{aligned}$$
Since $f\in C^1(\Bbb R),$ so is the composition $f\circ\pi^2,$ where $\pi:(a,z)\mapsto z.$ Therefore, $\frac{\partial (f\circ\pi^2)}{\partial x}$ exists and is continuous, so we can apply the above corollary to functions $G$ and $H.$
I think $$\begin{aligned}\frac{\partial F_3}{\partial y}(x,y)&=G'(y)-G'(x)=f(z^2)\cdot 0-f(z^2)\cdot 0+\int_0^y\underbrace{\frac{\partial (f\circ\pi^2)}{\partial x}}_{=0}-f(z^2)\cdot 0+f(z^2)\cdot 0-\int_0^x\underbrace{\frac{\partial(f\circ\pi^2)}{\partial x}}_{=0}\\&=0\end{aligned}$$
and analogously $$\frac{\partial F_3}{\partial x}(x,y)=H'(x)-H'(y)=0,$$ so $\operatorname{grad} F_3(x,y)=(0,0)$ and $\nabla F(x,y)=\begin{bmatrix}1&0\\0&1\\0&0\end{bmatrix}$ but I'm not so sure about my result.
Question:

Is there anything wrong and how can we formalize the answer?

 A: First and foremost, I applaud your attempt and the presentation of the content in your post. In particular, it is highly beneficial that you have stated the theorems you need to use as-is with all definitions and conditions explicit ; it assures me that I don't need to use external sources to aid you in the completion of your problem and the analysis of your attempt.

Now, let's go through your work.
1

Let $G,H : \mathbb R \to \mathbb R, G(t) = F_3(0,t), H(t) = F_3(t,0)$.

While I understand the reason behind defining these functions is perhaps to reflect the simultaneously changing upper and lower limit in the integral defining $F_3$ , you can already see that $$
G(t) = F_3(0,t) = \int_0^t f(z^2)dz  = -\int_t^0 f(z^2)dz = -F_3(t,0) = -H(t)
$$
that is, $G = -H$ as functions of $t$. This observation would have helped you reduce some computational effort.
2

The computations $F_3(x,y) = G(y) -G(x)$ and $F_3(x,y) = H(x) - H(y)$.

These are performed correctly, and recognize that we could have avoided one of these with the shortcut we used in section $1$ above.
3

"Since $f\in C^1(\mathbb R)$ ... to functions $G$ and $H$".

I'm not very sure about this step, let's dissect it. First of all, which corollary are you trying to use? You're trying to prove that $G$ and $H$ are
differentiable functions. Theorem 11.1 is not useful because it has limits which don't change, so it's not the corollary you're set up to use here.
If you're using Theorem 11.2, then you need to carefully see what fits where. For example, let's try fitting the corollary to $G$. Compare what the corollary wants to what we have : $$
F(y) = \int_{u(y)}^{v(y)} f(x,y)dx \leftrightarrow G(t) = \int_0^t f(z^2)dz
$$
Therefore, matching the lower and upper limits, you want $u(t) = 0$ and $v(t) = t$. What about the inner term? You have $f(x,y)dx$ on the LHS and $f(z^2)dz$ on the RHS. Now I can see where the $\pi$ comes in : you're basically trying to create the function $f(z^2)$ from a two-dimensional function, right? So define $g(z,t)$ by $g(z,t) = f(z^2)$, and you want to apply the corollary to the functions $g(z,t) = f(z^2), u(t) = 0, v(t)=t$.
Which is why I was unsure why you wanted $\pi$ to go from $(a,z) \to z$ : you want to prove that $g$ is continuous, using the fact that $f$ is continuous, but a projection usually takes you in the other direction? In short, I don't know what that part has exactly achieved.
There are also some missing details, because one has to do some work to show that $g$ is continuous.
What I would write in this situation is (sans some details that I would provide if I were e.g. in an exam):

We wish to prove that $G$ and $H$ are differentiable functions. To do this, we note that $$
G(t) = \int_0^t f(z^2)dz = \int_{u(t)}^{v(t)} g(z,t)dz
$$
where $$
u(t) = 0 \quad ; \quad v(t) =t \quad ; \quad g(z,t) = f(z^2)
$$
Note that $g$ is continuous. To prove this, observe that the function $f(z^2)$ as a one-dimensional function of $z$ is continuous using composition rules. Then, given any $\epsilon>0$, observe that $g(z,t) - g(z',t') = f(z^2)-f(z'^2)$, therefore using the $\delta$ obtained from using $\epsilon$ in the continuity definition of $f(z^2)$ gives that $d((z,t) ,(z',t'))<\delta$ implies that $|g(z,t)-g(z',t')|<\epsilon$. Thus $g$ is continuous.
The constancy of $t$ in the expression for $g(z,t)$ yields easily that $\frac{\partial}{\partial t}g(z,t) = 0$. Therefore, the corollary applies and $G,H$ are differentiable.

4
This is a very important point that you've missed.

You know that $G,H$ are one-dimensional differentiable functions. How did you conclude that $F_3(x,y)$ was differentiable from here?

Note that you will probably be able to prove that $F_3$ has partial derivatives without trouble. However, proving that $F_3$ is differentiable is stronger than asserting that it has partial derivatives : and that's part of the exercise.
How did you transition from one-dimensional to two-dimensional differentiability here? It's not trivial enough to be left out as a detail, I think.
The answer to that is to define the appropriate two-dimensional analogues $G_,(x,y) = G(x)$ and $H_,(x,y) = H(y)$, so that $F_3(x,y) = G_,(x,y) + H_,(x,y)$. Are these two differentiable? Indeed, thankfully they are (it is trivial to check this by the definition of multivariable differentiability, and one can find the partial derivatives from there as well). Therefore, $F_3$ is differentiable, as a difference of 2-dimensional differentiable functions. This is important, because you need this to find the derivatives of $F_3$.
Also, this shows that $F$ is differentiable since it is component-wise differentiable.
5
The computations for $\frac{\partial F_3}{\partial x}$ and $\frac{\partial F_3}{\partial y}$ are unfortunately incorrect. Indeed, I'm not sure how the jump from $\frac{\partial F_3}{\partial y} = G'(y) - G'(x)$ was made : it is a bit casual to make such a jump because one actually needs to apply corollary $11.2$, and I don't see how it was applied in the given situation.
Of course, the computations for $\frac{\partial F_1}{\partial x}\frac{\partial F_2}{\partial x}, \frac{\partial F_1}{\partial y},\frac{\partial F_2}{\partial y}$ are correct.
We start with $$
\frac{\partial F_3}{\partial x}(x,y) =\frac{\partial G_,}{\partial x}(x,y)  + \frac{\partial H_,}{\partial x}(x,y)
$$
Now, $H_,(x,y) = H(y)$ doesn't depend upon $x$ so that term is $0$. We are only left with $\frac{\partial G_,}{\partial x}(x,y)$, which we can check is equal to $G'(x)$ rather easily. Finally, by corollary $11.2$, we can get $$
G'(x) = g(v(x),x)v'(x) - g(u(x),x)u'(x) + \int_{u(x)}^{v(x)} \frac{\partial g}{\partial t}(z,t) dz
$$
which , because of what we already have, leads to $$
g(v(x),x)v'(x) = f(x^2) \\
g(u(x),x)u'(x) = 0 \\
\int_{u(x)}^{v(x)} \frac{\partial g}{\partial t}(z,t) dz = 0
$$
So you just get $G'(x) = f(x^2) = \frac{\partial F_3}{\partial x}(x,y)$.
In quite similar fashion, we note that $$
\frac{\partial F_3}{\partial y}(x,y) = \frac{\partial G_,}{\partial y}(x,y) +\frac{\partial H_,}{\partial y}(x,y) 
$$
The derivative of $G_,$ vanishes and that of $H_,$ is , rather analogously, just $-f(y^2)$. Therefore, the final answer is $$
\nabla f(x,y) = \begin{pmatrix}
1 & 0 \\
0 & 1 \\
f(x^2) & -f(y^2)
\end{pmatrix}
$$
