Proof that F'(z)=$\frac{∂F}{∂x}$=$\frac{1}{i}\frac{∂F}{∂y}$

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Let $f$ be continuous on an open connected subset $Ω$ and let $f = F'$ on $Ω$.

Let $γ$ be a smooth curve in $Ω$ with $γ(0) = a$ and $γ(1) = b$, prove that

$$F'(z)=\frac{∂F}{∂x}=\frac{1}{i}\frac{∂F}{∂y}$$

and use that to prove

$$\int_γ f \;\mathrm{dz}=F(b)-F(a)$$


For the first part, I am unsure how to even start proving this.


Regarding the second part.

If F'(z)=f(z), then is F'($\gamma(t)$)=f($\gamma(t)$)? And thus $$\int_γ f \;\mathrm{dz}=\int_0^1 f(\gamma(t))\gamma'(t)dt =\int_0^1 F'(\gamma(t))\gamma'(t)dt$$$$=F(\gamma(1))-F(\gamma(0))$$

It kinda feels wrong to just substitute f($\gamma(t)$) with F'($\gamma(t)$).