Consider the following functions.

$w=y^3-6x^{2}y$

$x=e^s,y=e^{t}x$ and $y.\frac{\text{d}\text{e}\text{l}\text{w}}{\text{d}\text{e}\text{l}\text{x}}=$ Amazing job!

$\frac{\text{d}\text{e}\text{l}\text{w}}{\text{d}\text{e}\text{l}\text{y}}=$ Great! $s$ and $t.\frac{\text{d}\text{e}\text{l}\text{x}}{\text{d}\text{e}\text{l}\text{s}}=$ Impressive work!

$\frac{\text{d}\text{e}\text{l}\text{y}}{\text{d}\text{e}\text{l}\text{s}}=$ Good work.

$\frac{\text{d}\text{e}\text{l}\text{x}}{\text{d}\text{e}\text{l}\text{t}}=$ Well done.

$\frac{\text{d}\text{e}\text{l}\text{y}}{\text{d}\text{e}\text{l}\text{t}}=$ That's $it!$

Find $\frac{\text{d}\text{e}\text{l}\text{w}}{\text{d}\text{e}\text{l}\text{s}}$ and $\frac{\text{d}\text{e}\text{l}\text{w}}{\text{d}\text{e}\text{l}\text{t}}s$ and $t.\frac{\text{d}\text{e}\text{l}\text{w}}{\text{d}\text{e}\text{l}\text{s}}=$

$\frac{\text{d}\text{e}\text{l}\text{w}}{\text{d}\text{e}\text{l}\text{t}}=$

Check which variable$(s)$ should $bein$ your answer.

Evaluate $\frac{\text{d}\text{e}\text{l}\text{w}}{\text{d}\text{e}\text{l}\text{s}}$ and $\frac{\text{d}\text{e}\text{l}\text{w}}{\text{d}\text{e}\text{l}\text{t}}$ when $s=-2$ and $t=4.$

$\frac{\text{d}\text{e}\text{l}\text{w}}{\text{d}\text{e}\text{l}\text{s}}=,T^{\text{T}\text{h}\text{a}\text{t}\text{'}\text{s}it!}$

$\frac{\text{d}\text{e}\text{l}\text{w}}{\text{d}\text{e}\text{l}\text{t}}=,x$

Enter $an$ exact answer with $no$ rounding.