# $yz\,dx-2xz\,dy+(xy-y^3z)\,dz=0$

My attempt at the question

What I don't know is do we integrate partially w.r.t. z to find the value of Φ (z) here?

$$\require{begingroup} \begingroup \newcommand{\dd}{\;\mathrm{d}}$$ $$yz\dd x-2xz \dd y + (xy-y^3z) \dd z=0.\tag{1}$$ $$P=yz$$, $$Q=-2xz$$ and $$R=xy-y^3z$$
Now \begin{align*} &P\left(\frac{\partial Q}{\partial z}-\frac{\partial R}{\partial y}\right)+ Q\left(\frac{\partial R}{\partial x}-\frac{\partial P}{\partial z}\right)+ R\left(\frac{\partial P}{\partial y}-\frac{\partial Q}{\partial z}\right)\\[.8em] &=P(-2x-x+3y^2z)+Q(y-y)+R(z+2z)\\ &=yz(-3x+3y^2z)+(-2xz)(0)+(xy-y^3z)(z+2z)\\ &=-3xzy+3y^3z^2+0+3xyz-3y^3z^2\\ &=0 \end{align*} Hence equation $$(1)$$ is integrable.
Let $$z$$ be a constant in $$(1)$$, then $$\dd z=0$$. Integrating both sides \begin{align*} \int\frac1{2xz}\dd x-\int\frac1{2yz}\dd y&=0\\ \frac1{2z}\ln x - \frac1z \ln y &= const\\ \frac{x^{1/2}}y &= \Phi(z) \tag{2} \end{align*} Differentiating w.r.t $$x$$, $$y$$, $$z$$ $$\frac{x^{-1/2}}{2y} \dd x + \left(-\frac{x^{1/2}}{y^2}\right)\dd y+(-\Phi'(z)) \dd z = 0 \tag{3}$$ From equations $$(1)$$ and $$(3)$$ $$\frac{yz}{1/(2x^{1/2}y)}=\frac{-2xz}{-x^{1/2}/y^2}=\frac{xy-y^3z}{-\Phi'(z)}.$$ $$\endgroup$$

You need to find an integrating factor. If you treat this problem like a puzzle you can combine like terms to find $$y\,d(xz)-2(xz)\,dy=y^3zdz$$ and from that that you get an integrable expression by dividing by $$y^3$$, that is, the integrating factor should turn out to be $$y^{-3}$$.
In your solution, I'd have avoided the square root to get $$xy^{-2}=\Phi(z)$$ as the equation for the solution surface. Then the derivative of that is $$y^{-2}\,dx-2xy^{-3}\,dy =\Phi'(z)\,dz\\ yz\,dx - 2xz\,dy = y^3z\Phi'(z)\,dz\\ \implies y^3z\Phi'(z)\,dz=(y^3z-xy)\,dz\implies \Phi'(z)=1-z^{-1}\Phi(z)$$ which can now be easily solved.
In your version with $$Φ(z)=\frac{x^{1/2}}y$$, pick one of the relations and eliminate $$y$$ against $$Φ$$. The first two fractions are equal after simplification, the equality to the last term gives $$2x^{1/2}y^2z=y^3\frac{Φ(z)^2-z}{-Φ'(z)}\iff 2zΦ'(z)Φ(z)=z-Φ(z)^2\implies zΦ(z)^2=\frac12z^2+C$$