第一问（1）
using TySymbolicMath
@variables x y z
f1=2xytan(x)+exp(3xy)+cos(yz)

##对x求导
df_dx=derivative(f1,x)
##对y求导
df_dy=derivative(f1,y)
##对z求导
df_dz=derivative(f1,z)

得到答案：
df_dx=2ytan(x) + 3yexp(3xy) + 2xy*(1 + tan(x)^2)
df_dy=2xtan(x) + 3xexp(3xy) - zsin(yz)
df_dz=-ysin(y*z)

第一问（2）
using TySymbolicMath
@variables x y z
g=(x+y+yz)/sin(xy)

##对x求导
df_dx=derivative(g,x)
##对y求导
df_dy=derivative(g,y)
##对z求导
df_dz=derivative(g,z)

得到答案：
df_dx=1 / sin(xy) - ycos(xy)((x + y + yz) / (sin(xy)^2))
df_dy=(1 + z) / sin(xy) - xcos(xy)((x + y + yz) / (sin(xy)^2))
df_dz=y / sin(x*y)

第二问
using TySymbolicMath
@variables x y
t=100exp(-x^2-2y^2)

dt_dx=derivative(t, x)
dt_dy=derivative(t, y)

g = TySymbolicMath.gradient(t, [x, y])
x0=0.5
y0=-0.5
grad_at_point=[substitute(gi, Dict(x=>x0, y=>y0)) for gi in g]
grad_x=grad_at_point[1]
grad_y=grad_at_point[2]
v=[3,4]
v_x=sqrt(v[1]^2+v[2]^2)
u=v/v_x
directional_deriv=grad_xu[1]+grad_yu[2]

得到梯度向量：[-47.236655274101466，94.47331054820293]
方向导数：94.47331054820293