Let $g:\mathbb{R}^3 \to \mathbb{R}^2$ be given by $g((x,y,z)) = (g_1(x,y,z)-11, g_2(x,y,z)-3)$. The feasible set is described by $g^{-1}(\{0\})$, and you are trying to find $z$ as a function of $x,y$ locally.
Let me abuse notation and write $g(p,z)$ with $p=(x,y)$ to simplify my life.
Then you need to show that ${\partial g(p,z)) \over \partial p}$ is invertible at a solution $(\hat{p}, \hat{z})$, and the implicit function theorem gives some $\pi$ defined locally around $\hat{z}$ such that $g(\pi(z),z) =0$ for
$z$ near $\hat{z}$, and we have
${\partial \pi(\hat{z})) \over \partial z} = - {\partial g(\hat{p},\hat{z})) \over \partial p}^{-1} {\partial g(\hat{p},\hat{z})) \over \partial z}$.
If we let $\phi(z) = f(\pi(z),z)$ (abusing notation again), then at a solution we will have ${\partial \phi(\hat{z})) \over \partial z} = 0$, using the
chain rule this reduces to
${\partial f(\hat{z},\hat{p})) \over \partial z} = {\partial f(\hat{z},\hat{p})) \over \partial p} {\partial g(\hat{p},\hat{z})) \over \partial p}^{-1} {\partial g(\hat{p},\hat{z})) \over \partial z}$.
(As an aside, when used as a method for numerical optimization, this is known as 'reduced gradients'.)
The first thing to remember is that the Fourier transform $\mathcal{F}$ is linear: $$
\mathcal{F}(\alpha f+\beta g)=\alpha \mathcal{F}(f)+\beta \mathcal{F}(g)
$$
It also changes differentiation into multiplication. Let $F(\omega)$ be the transform of $f(t)$. Then:
$$
\mathcal{F}(D_tf)(\omega)=\int_{-\infty}^{\infty}D_tf(t)e^{-j\omega t}\mathrm{d}t = f(t)e^{-j\omega t}|_{-\infty}^{\infty}+j\omega\int_{-\infty}^{\infty} f(t)e^{-j\omega t}\mathrm{d}t = 0 + j\omega F(\omega)
$$
For this to actually be correct, the boundary term from integration by parts has to disappear in the limit. This imposes some technical conditions. Also, if $f$ happens to be periodic, the Fourier transform (likely) won't exist in the standard sense and may require the use of distributions. I won't dwell on these issues here (mostly because I don't actually know a whole lot about all the theory behind these concerns), but perhaps you can make a separate question specifically about when this process works, or maybe someone else can add another answer regarding this.
Anyway, I'll just do the formal manipulations here (as is usually done in physics classes). In your case, we have functions of several variables $({\bf r}, t)$. However, this isn't a problem. For example:
$$
\mathcal{F}(\partial_t{\bf e})({\bf r},\omega)=\int_{-\infty}^{\infty}\partial_t{\bf e}({\bf r},t)e^{-j\omega t}\mathrm{d}t= j\omega {\bf E}({\bf r},\omega)
$$
Here, the process is exactly the same as the above since we can keep ${\bf r}$ fixed and so you can use integration by parts on $\partial_t$. If you're worried about vector terms, notice that this is no different from integrating any other function ${\bf f}(u)$ with values in $\mathbb{R^3}$ (I'll omit the integration limits because they don't really matter):
$$
\int{\bf f}(u)\mathrm{d}u=\left(\int f_x(u)\mathrm{d}u,\int f_y(u)\mathrm{d}u,\int f_z(u)\mathrm{d}u\right)
$$
Or, in component notation ($i=x,y,z$):$$
\left(\int{\bf f}(u)\mathrm{d}u\right)_i=\int f_i(u)\mathrm{d}u$$
The curl is also not problematic as the spatial derivatives commmute with time integration:
$$
\int\partial_x{\bf f}({\bf r},t)\mathrm{d}t=\partial_x\int{\bf f}({\bf r},t)\mathrm{d}t
$$
Because of this, you can immediately conclude that $\mathcal{F}(\partial_x{\bf f})=\partial_x\mathcal{F}({\bf f})$. Writing out only the $x$-component:
$$
\left(\mathcal{F}(\nabla\times{\bf f})\right)_x=\mathcal{F}(\partial_yf_z-\partial_zf_y)=\partial_yF_z-\partial_zF_y = (\nabla\times{\bf F})_x
$$
Here ${\bf F}=\mathcal{F}({\bf f})$. The other two components work in exactly the same way. Finally, take the equation $$\nabla\times{\bf h}={\bf j}+\epsilon_0\partial_t{\bf e}+\partial_t{\bf p}$$
and apply $\mathcal{F}$ to both sides. By linearity and what we've said about $\nabla\times$, we find
$$
\mathcal{F}(\nabla\times{\bf h})=\mathcal{F}({\bf j}+\epsilon_0\partial_t{\bf e}+\partial_t{\bf p})$$
$$
\nabla\times\mathcal{F}({\bf h})=\mathcal{F}({\bf j})+\epsilon_0\mathcal{F}(\partial_t{\bf e})+\mathcal{F}(\partial_t{\bf p})
$$
Keeping in mind what $\mathcal{F}$ does to $\partial_t$, we finally get:
$$
\nabla\times{\bf H}={\bf J}+j\omega\epsilon_0{\bf E}+j\omega{\bf P}
$$
The other equation follows suit.
As one commenter pointed out, this is a Fourier transform on time. It is also possible to do a Fourier transform on the spatial coordinates. These two approaches are complementary; often, both Fourier transforms are taken and so we change from $({\bf r},t)$-space to $({\bf k},\omega)$-space where ${\bf k}$ is the wave vector.
Best Answer
$\textbf{Hints:}$ Check out the Leibniz Integral Rule to handle cases where you're differentiating with respect to a variable different than the one being integrated and apply the Fundamental Theorem of Calculus to cases where you're differentiating with respect to the same variable that is being integrated.