I had a question related to the properties of partial derivatives and those are used casually but aren't mentioned explicitly in any books the way those properties are mentioned in case of ordinary derivatives.
1.Is $\frac{\partial y}{\partial x}$ the inverse of $\frac{\partial x}{\partial y}$? For example if $z=x^2y$ then $\frac{\partial z}{\partial x}=2xy$. Again,$x=\sqrt{\frac{z}{y}}$,so $\frac{\partial x}{\partial z}=\frac{1}{2\sqrt{\frac{z}{y}}} \times \frac{1}{y}=\frac{1}{2xy}$. So,in this random example the property does seem to be true. I tried for few other examples as well.Or are there any examples where this property fails?
2.Is $\frac{\partial y}{\partial x}=\frac{\partial y}{\partial t} \times \frac{\partial t}{\partial x}$? For example,if $z=\sin(5xy+y)$,then $\frac{\partial z}{\partial x}$ is done as $5y\cos(5xy+y)$. Here,the CHAIN RULE of ordinary derivatives has been applied,as it is the same as $\frac{\partial \sin(5xy+y)}{\partial (5xy+y)} \times \frac{\partial (5xy+y)}{\partial x}$. So indeed the chain rule of ordinary derivatives applies here but it is not acknowledged as the chain rule for multivariable calculus as its chain rule is different which is $\frac{dz}{dt}=\frac{\partial z}{\partial x}\frac{dx}{dt}+\frac{\partial z}{\partial y}\frac{dy}{dt}$.
Among all the examples I have encountered so far,the properties of ordinary derivatives seem to fit in partial derivatives. But these aren't mentioned properly in the books so I assumed these to be not true in general. But these properties are used in partial differentiation(the no. $2$ mostly which I mentioned as without the chain rule of ordinary derivatives,there is no other way of partially differentiating $\sin$ or $\cos$ functions). So I want to know if these properties are valid,why aren't they mentioned anywhere. Or are the valid in general?
As with the vast majority of questions about partial derivatives, the trouble has to do with our suboptimal notation for them that usually doesn’t indicate what’s being held fixed. To clear things up a bit, I’ll use the notation $\left.\frac{\partial f}{\partial x}\right|_y$ to indicate that the variable $y$ is being held fixed while we differentiate $f$ with respect to $x$.
In your first example, you found that $\left.\frac{\partial z}{\partial x}\right|_y=\left(\left.\frac{\partial x}{\partial z}\right|_y\right)^{-1}$. This isn’t very suprising – you treated $y$ as a constant (so it might as well have been $2$ or $\pi$ instead), and then this relationship follows directly from the corresponding relationship for ordinary derivatives. Ordinary derivatives are just partial derivatives of univariate functions; partial derivatives are just ordinary derivatives of multivariate functions interpreted as parametrized families of univariate functions.
Somewhat more formally, we could say that we have a bivariate function $z(x,y)$; we consider it as a parametrized family of univariate functions, $z_y(x)=z(x,y)$; we invert those functions to obtain another parametrized family of univariate functions, $x_y(z)$; and then we consider this family as one bivariate function, $x(z,y)=x_y(z)$; so we have
$$ \left.\frac{\partial z(x,y)}{\partial x}\right|_y:=\frac{\mathrm dz_y(x)}{\mathrm dx}=\left(\frac{\mathrm dx_y(z)}{\mathrm dz}\right)^{-1}=: \left.\frac{\partial x(z,y)}{\partial z}\right|_y\;. $$
That’s just a roundabout formal way of saying “treat $y$ as a constant”.
The counterexample given by Ted Shifrin concerns a different case. Here we have two bivariate functions, $x(r,\theta)=r\cos\theta$ and $y(r,\theta)=r\sin\theta$, and we invert them as a whole, not holding anything fixed, not going through an imaginary stage of univariate functions: $r(x,y)=\sqrt{x^2+y^2}$, $\theta(x,y)=\operatorname{atan2}(y,x)$. Here there’s no reason to think there should be any particular relationship between $\left.\frac{\partial r}{\partial x}\right|_y$ and $\left.\frac{\partial x}{\partial r}\right|_\theta$. Note that here we have two different variables being held fixed, $y$ and $\theta$, whereas in what you did it was the same variable $y$ being held fixed in both cases.
If we do hold the same variable fixed, we get the same result here as you did: $\left.\frac{\partial x}{\partial r}\right|_\theta=\cos\theta$ and, with $r=\frac x{\cos\theta}$, $\left.\frac{\partial r}{\partial x}\right|_\theta=\frac1{\cos\theta}$, and also $\left.\frac{\partial r}{\partial x}\right|_y=\frac x{\sqrt{x^2+y^2}}$ and, with $x=\sqrt{r^2-y^2}$, $\left.\frac{\partial x}{\partial r}\right|_y=\frac r{\sqrt{r^2-y^2}}=\frac{\sqrt{x^2+y^2}}x$.
The same applies to what you wrote about the chain rule. Again, you were keeping $y$ fixed in all of what you did, so it might as well have been $2$ or $\pi$, so it’s not surprising that everything works as for univariate functions: $\left.\frac{\partial z}{\partial x}\right|_y=\left.\frac{\partial z}{\partial t}\right|_y\cdot\left.\frac{\partial t}{\partial x}\right|_y$. As a general principle, you can take any result for ordinary derivatives and tack on any number of variables that you hold fixed to obtain a corresponding result for partial derivatives, as long as you use the same set of fixed variables for each derivative.