IT’S VERY EASY TO MAKE A MISTAKE IN THESE CALCULATIONS, SO PLEASE CHECK MY WORKING BELOW!
z=30x²+100xy-3y².
z=30cosh²(7t)cos²(7s)
+100(cosh(7t)cos(7s))(sinh(7t)sin(7s))
-3sinh²(7t)sin²(7s).
sin(14s)=2sin(7s)cos(7s) (trig identity).
sinh(14t)=2sinh(7t)cosh(7t) (hyp trig identity).
So z=30cosh²(7t)cos²(7s)+25cosh(14t)sin(14s)-3sinh²(7t)sin²(7s).
∂z/∂t=30cos²(7s)(14cosh(7t)sinh(7t))
+25sin(14s)(14sinh(14t))
-3sin²(7s)(14sinh(7t)cosh(7t)).
∂z/∂t=210cos²(7s)sinh(14t)+350sin(14s)sinh(14t)-21sin²(7s)sinh(14t),
∂z/∂t|(ln(5)/14,π/7)=210sinh(ln(5))+0-0.
sinh(ln(5))=½(5-1/5)=12/5, so ∂z/∂t|(ln(5)/14,π/7)=210(12/5)=504.
[Also:
cosh(ln(√5))=½(e^(½ln(5))+e^(-½ln(5)))=
½(√5+(1/√5))=
½(5+1)/√5=3/√5=3√5/5.
sinh(ln(√5))=½(e^(½ln(5))-e^(-½ln(5)))=
½(√5-(1/√5))=
½(5-1)/√5=2/√5=2√5/5.
∂z/∂t=(∂z/∂x)(∂x/∂t)=60x(∂x/∂t)+100y(∂x/∂t)=
(∂x/∂t)(60x+100y)=(-14√5/5)(60(-3√5/5)+0)=504.
But note that some partial differentials become zero when substituting for t and s. This is why I decided to opt for substituting for x and y at the beginning so that z was in terms of s and t.]
∂z/∂s=-210cosh²(7t)sin(14s)+350cosh(14t)cos(14s)-3sinh²(7t)(14sin(7s)cos(7s)),
∂z/∂s=-210cosh²(7t)sin(14s)+350cosh(14t)cos(14s)-21sinh²(7t)sin(14s).
∂z/∂s|(ln(5)/14,π/7)=0+350cosh(ln(5))-0.
cosh(ln(5))=½(5+1/5)=13/5, so ∂z/∂s|(ln(5)/14,π/7)=350(13/5)=910.
Even if I made an error, I hope you can see the method clearly.