$$f(x,y,z)= \int_{0}^{x}t^tdt + x ^{\sin(y^z)}$$
The derivative would be $f'(x,y,z)(h^1,h^2,h^3)= \frac{\partial f}{\partial x}h^1+\frac{\partial f}{\partial y}h^2+\frac{\partial f}{\partial z}h^3.$
$\frac{\partial f}{\partial x}=...?$ Could someone put in a step by step on there? $\frac{\partial f}{\partial y}=x^{\sin(y^z)}\ln( \sin (y^z))\cos (y^z)zy^{z-1}$ $\frac{\partial f}{\partial z}=x^{\sin(y^z)}\ln( \sin (y^z))\cos (y^z)y^z \ln z ?$
Since $t\mapsto t^t$ is smooth, the fundamental theorem of calculus tells us that $$\frac{\partial}{\partial x}\left(x\mapsto\int_0^xt^t\mathrm{d}t\right)(x)=x^x.$$
Next, recall that $$y^z=\mathrm{e}^{z\ln y},\quad x^{\sin\left(y^z\right)}=\mathrm{e}^{\sin\left(y^z\right)\ln x}$$ so that $$\frac{\partial}{\partial y}\left(y\mapsto y^z\right)(y)=\mathrm{e}^{z\ln y}\frac{z}{y}=zy^{z-1}$$ and then $$\frac{\partial}{\partial y}\left(y\mapsto x^{\sin\left(y^z\right)}\right)(y)=\mathrm{e}^{\sin\left(y^z\right)\ln x}\cos\left(y^z\right)zy^{z-1}\ln x=x^{\sin\left(y^z\right)}\cos\left(y^z\right)zy^{z-1}\ln x.$$
Finally, we have $$\frac{\partial}{\partial z}\left(y\mapsto x^{\sin\left(y^z\right)}\right)(z)=x^{\sin\left(y^z\right)}\cos\left(y^z\right)\ln y\ln x.$$