Chain rule partial derivative 
Find the partial derivatives $\partial Z / \partial u $ and $\partial Z / \partial \nu$ for the following function:
  $$
Z(x,y,z) = 2x^3 - 3xy^2 + 0.75~yu - 5u^2\quad \text{where}~x = \sqrt{u+\nu}~\text{and}~y=\nu^2
$$

No matter how hard I try I couldn't solve it and couldn't find anything similar on internet either. Could you take a look at it?

We calculate  $\frac{\partial Z}{\partial  u}$ in two ways.
First variant: We consider
\begin{align*}
Z(x(u,v),y(u,v),z(u,v))=2x^3-3xy^2+\frac{3}{4}yz-5z^2
\end{align*}
where
\begin{align*}
x=\sqrt{u+v},\qquad y=v^2,\qquad z=u
\end{align*}

According to the chain rule we obtain
\begin{align*}
\color{blue}{\frac{\partial   Z}{\partial  u}}
&=\frac{\partial  Z}{\partial x}\,\frac{\partial x}{\partial u}
+\frac{\partial  Z}{\partial y}\,\frac{\partial y}{\partial u}
+\frac{\partial  Z}{\partial z}\,\frac{\partial z}{\partial u}\\
&=\left(6x^2-3y^2\right)\frac{1}{2}\left(u+v\right)^{-\frac{1}{2}}+
\left(-6xy+\frac{3}{4}u\right)\cdot0+\left(\frac{3}{4}y-10z\right)\cdot 1\\
&=\left(6(u+v)-3v^4\right)\frac{1}{2}\left(u+v\right)^{-\frac{1}{2}}+\left(\frac{3}{4}v^2-10u\right)\\
&\,\,\color{blue}{=3\left(u+v\right)^{\frac{1}{2}}-\frac{3}{2}v^4\left(u+v\right)^{-\frac{1}{2}}+\frac{3}{4}v^2-10u}\tag{1}
\end{align*}

Second variant:  We consider
\begin{align*}
Z(u,v)&=\left.  2x^3-3xy^2+\frac{3}{4}yz-5z^2\right|_{x=\sqrt{u+v}, y=v^2, z=u}\\
&=2\left(u+v\right)^{\frac{3}{2}}-3\left(u+v\right)^{\frac{1}{2}}v^4+\frac{3}{4}v^2u-5u^2\tag{2}
\end{align*}

We obtain from (2)
\begin{align*}
\color{blue}{\frac{\partial   Z}{\partial  u}}
&=\frac{\partial}{\partial u}\left(2\left(u+v\right)^{\frac{3}{2}}-3\left(u+v\right)^{\frac{1}{2}}v^4+\frac{3}{4}v^2u-5u^2\right)\\
&\,\,\color{blue}{=3\left(u+v\right)^{\frac{1}{2}}-\frac{3}{2}v^4\left(u+v\right)^{-\frac{1}{2}}+\frac{3}{4}v^2-10u}\tag{3}
\end{align*}

We observe (1) and (3) are equal as expected. The partial derivative $\frac{\partial Z}{\partial v}$ can be calculated similarly.
