Problem:
Given $x=u+v$, $y=u^2+v^2$, $z=u^3+v^3$ ,
Find $\frac{δz}{δx}$ and $\frac {δz}{δy}$.
Now I started off this problem by calculating dx, dy, and dz in terms of u and v from the given equations. But i do not know what to do next. Can someone please help me? TIA :)
Just a quick note on notation: you use $\mathrm{d}$ (the 'straight $d$') when the function that you're differentiating only has one variable, and $\partial$ (the 'curly $d$' which is rendered with
\partial) when the function that you're differentiating has more than one variable.Thus, it would be appropriate to write $\partial x$ in this case, and the same for $\partial y$ and $\partial z$.
Now, the chain rule is super simple, and hopefully can be easy to remember: if $f$ is a function of $x$, and $x$ is a function of $u$, then the chain rule is $$ \frac{\mathrm{d}f}{\mathrm{d}u} = \frac{\mathrm{d}f}{\mathrm{d}x}\frac{\mathrm{d}x}{\mathrm{d}u}. $$ You introduce $\mathrm{d}x/\mathrm{d}x$ (which is just $1$, so is totally valid) that you group in a particular way so that you can differentiate two things which are easier. Notice that each of $f$ and $x$ are a function in one variable, hence the straight $\mathrm{d}$'s (though some authors might still write $\partial f/\partial x$ to emphasise that $f$ is a function of $x$, but that $x$ is also a function of something else).
If, however, the function $f$ was a function of the variables $x_{1}$, $x_{2}$, and $x_{3}$, with each of $x_{1}$, $x_{2}$, and $x_{3}$ functions of $u$, then $$ \frac{\mathrm{d}f}{\mathrm{d}u} = \frac{\partial f}{\partial x_{1}}\frac{\mathrm{d} x_{1}}{\mathrm{d} u} + \frac{\partial f}{\partial x_{2}}\frac{\mathrm{d} x_{2}}{\mathrm{d} u} + \frac{\partial f}{\partial x_{3}}\frac{\mathrm{d} x_{3}}{\mathrm{d} u}. $$ We just sum the derivatives of $f$ with respect to the other variables, and use the chain rule in each case. Observe the use of straight $\mathrm{d}$'s and curly $\partial$'s. I hope it is clear how this extends if $f$ is a function of many other variables.
For your particular example, you have $$ x=u+v, \hspace{20pt} y=u^2+v^2, \hspace{20pt} z=u^3+v^3. $$ Each of $x$, $y$, and $z$ are functions of two variables, so we use the curly $\partial$ in their derivatives. You need to calculate $$ \frac{\partial x}{\partial u},\hspace{20pt} \frac{\partial x}{\partial v},\hspace{20pt} \frac{\partial y}{\partial u},\hspace{20pt} \frac{\partial y}{\partial v},\hspace{20pt} \frac{\partial z}{\partial u},\hspace{20pt} \frac{\partial z}{\partial v}, $$ which are straightforward (and I shall leave to you). Then you want $$ \frac{\partial z}{\partial x} \hspace{20pt}\text{and}\hspace{20pt} \frac{\partial y}{\partial x}. $$ Expand these using the multivariate chain rule to get $$ \frac{\partial z}{\partial x} = \frac{\partial z}{\partial u}\frac{\partial u}{\partial x} + \frac{\partial z}{\partial v}\frac{\partial v}{\partial x} $$ and $$ \frac{\partial y}{\partial x} = \frac{\partial y}{\partial u}\frac{\partial u}{\partial x} + \frac{\partial y}{\partial v}\frac{\partial v}{\partial x}. $$ Notice that $\partial u/\partial x$ is just the reciprocal (one over) of what you calculated for $\partial x/\partial u$ (and analogously for the other like terms), so you have all the terms that you need to find these derivatives now.